{"id":1089,"date":"2015-08-02T14:08:16","date_gmt":"2015-08-02T14:08:16","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1089"},"modified":"2024-08-23T12:31:41","modified_gmt":"2024-08-23T12:31:41","slug":"resolucao-dos-exercicios-de-vestibulares-sobre-lancamento-obliquo","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/lancamento-obliquo\/resolucao-dos-exercicios-de-vestibulares-sobre-lancamento-obliquo\/","title":{"rendered":"Lan\u00e7amento obl\u00edquo – Resolu\u00e7\u00e3o dos exerc\u00edcios de vestibulares"},"content":{"rendered":"

Resolu\u00e7\u00e3o dos exerc\u00edcios de vestibulares sobre lan\u00e7amento obl\u00edquo<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

01-\u00a0O tempo que a bola permanece no ar est\u00e1 relacionado\u00a0 com a altura\u00a0 —\u00a0 maior altura, maior tempo de perman\u00eancia no ar\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

02-\u00a0I \u2013 Falsa \u2013 a acelera\u00e7\u00e3o \u00e9 constante e \u00e9 a acelera\u00e7\u00e3o da gravidade\u00a0\"\", sempre com dire\u00e7\u00e3o vertical e sentido para baixo.<\/b><\/span><\/span><\/span><\/p>\n

II \u2013 Correta \u2013 vide afirma\u00e7\u00e3o acima<\/b><\/span><\/span><\/span><\/p>\n

III \u2013 Falsa \u2013 no ponto mais alto da trajet\u00f3ria a velocidade \u00e9 m\u00ednima e vale V=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

03-\u00a0Se os dois \u00e2ngulos de lan\u00e7amento forem\u00a0complementares entre si (\u03b1<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ \u03b1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>), e a velocidade inicial for a mesma, (no caso, 20m\/so alcance horizontal \u00e9 o mesmo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

04-\u00a0a) Como a resist\u00eancia do ar \u00e9 desprezada, a velocidade horizontal inicial do proj\u00e9til \u00e9 constante e, em cada instante, a mesma do caminh\u00e3o. Assim, se ele partiu de um ponto P da carroceria do caminh\u00e3o, retornar\u00e1 ao mesmo ponto P e o deslocamento horizontal em rela\u00e7\u00e3o ao caminh\u00e3o ser\u00e1\u00a0zero.<\/b><\/span><\/span><\/span><\/p>\n

b) V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=20m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=80m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 6.400=400 + V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=77.5m\/s\u00a0 —tempo de subida\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0=77,5 \u2013 10t<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 t<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=7,75s\u00a0 —\u00a0 tempo que demora para subir e descer e se deslocar X na horizontal\u00a0 —\u00a0 t=2.7.75\u00a0 —\u00a0 t=15,5s\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.t=20,15,5\u00a0 —\u00a0\u00a0X=310m<\/b><\/span><\/span><\/span><\/p>\n

05-\u00a0(01) Falsa \u2013 \u00e9 o tempo de subida mais o tempo de descida<\/b><\/span><\/span><\/span><\/p>\n

(02) Verdadeira \u2013 veja (1)<\/b><\/span><\/span><\/span><\/p>\n

(04) Verdadeira \u2013 veja teoria – Se a altura maxima (h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>) \u00e9 igual ao alcance X\u00a0 —\u00a0 \u00a0tg\u03b1=4<\/b><\/span><\/span><\/span><\/p>\n

(08) E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 na altura m\u00e1xima V \u00e9 m\u00ednima, portanto E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0tamb\u00e9m ser\u00e1 m\u00ednima \u2013 Falsa<\/b><\/span><\/span><\/span><\/p>\n

(16) Falsa \u2013 como n\u00e3o existe atrito, o sistema \u00e9 conservativo e a energia mec\u00e2nica \u00e9 sempre a mesma em todos os pontos da trajet\u00f3ria<\/b><\/span><\/span><\/span><\/p>\n

Soma (02 + 04) = 06<\/b><\/span><\/span><\/span><\/p>\n

06-\u00a0Na altura m\u00e1xima a velocidade vetorial\u00a0\"\"\u00a0n\u00e3o \u00e9 nula, tem intensidade m\u00ednima e \u00e9 igual \u00e0 componente horizontal, ou seja,\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=20m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 20=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.1\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=40m\/s\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

07-\u00a0Na altura maxima\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=20m e t=4\/2\u00a0 —\u00a0 t=2s\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 20=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>.2 \u2013 10.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=20m\/s\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=20t \u2013 5t<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0No ponto mais alto\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=20\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=300\u00a0 —\u00a0 na altura m\u00e1xima h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2gh<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=300 \u2013 20h<\/b><\/span><\/span><\/span>m\u00e1qx<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=15m<\/b><\/span><\/span><\/span><\/p>\n

09-\u00a0a) Movimento na vertical\u00a0 —\u00a0 no ponto A de altura m\u00e1xima V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 YB) = Y(A) + V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>t \u2013 10t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 4,2 = 5,0 +0.t -5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 t=\u221a0,16\u00a0 —\u00a0\u00a0t=0,4s
\nb) Queda livre da altura Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5m\u00a0 — V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 quando chega ao solo Y=0\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t \u2013gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0=5 + 0t \u2013 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u00a0t=1s<\/b><\/span><\/span><\/span><\/p>\n

Sendo o choque el\u00e1stico, o tempo de subida \u00e9 igual ao tempo de descida\u00a0 —\u00a0\u00a0t=2s<\/b><\/span><\/span><\/span><\/p>\n

c) Movimento vertical\u00a0 —\u00a0 a batida na parede n\u00e3o afeta o tempo de queda (proje\u00e7\u00e3o na vertical) pois o choque \u00e9 el\u00e1stico —\u00a0 t=1s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 velocidade com que chega ao solo\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0 -10.1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=-10m\/s\u00a0 —\u00a0 se chega ao solo com velocidade de -10m\/s, sai do mesmo com velocidade de +10m\/s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No movimento horizontal ela demora t=0,4s para percorrer X=6m com velocidade constante V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0 6=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>.0,4\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=15m\/s\u00a0 —\u00a0 +15m\/s para a direita (movimento progressivo) e -15m\/s para a esquerda (movimento retr\u00f3grado)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

10-\u00a0V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos45<\/b><\/span><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=0,7V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 tempo que o dardo demora para para percorrer 16m na horizontal\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>cos45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.t\u00a0 —\u00a0 16=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.0,7.t\u00a0 —\u00a0 t=16\/0,7V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Na altura m\u00e1xima V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0 e t=16\/2.(0,7V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=16\/1,4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V\u00ad<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=0,7V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 10t\u00a0 —\u00a0 0=0,7V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 10.16\/1,4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=\u221a163,2=12,8m\/s\u00a0 —\u00a0 t=16\/0,7V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t=16\/12,8=1,8s\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

11-\u00a0a) A \u00fanica for\u00e7a que age sobre a bola (a resist\u00eancia do ar \u00e9 desprezada) durante todo o movimento \u00e9 a for\u00e7a peso, vertical e para<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

baixo.<\/b><\/span><\/span><\/span><\/p>\n

b) C\u00e1lculo do tempo que a bola demora\u00a0 a chegar at\u00e9 o goleiro percorrendo X=40m com velocidade horizontal constante e de valor V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=26.0,91\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=23,66m\/s\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 40=23,66.t\u00a0 —\u00a0 t=1,69s\u00a0 —\u00a0 c\u00e1lculo da altura, na dire\u00e7\u00e3o vertical, que a bola estar\u00e1 ao chegar ao goleiro nesse instante (t=1,69s)\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.t \u2013gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=26.0,42.1,69 \u2013 10.(1,69)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=18,45 \u2013 14,28\u00a0 —\u00a0 Y=4,17m\u00a0 —\u00a0 esse valor \u00e9 maior que 3m e assim, o\u00a0goleiro n\u00e3o consegue tocar a bola.<\/b><\/span><\/span><\/span><\/p>\n

c) C\u00e1lculo do tempo que a bola demora para chegar \u00e0 altura vertical de 1,5m\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1,5=10,92t \u2013 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0-10,92t + 1,5=0\u00a0 —\u00a0 \u0394=119,25 \u2013 30=89,25\u00a0 —\u00a0 \u221a\u0394=9,5\u00a0 —\u00a0 t=(10,92 \u00b19,5)\/2.5\u00a0 —\u00a0 considera-se o tempo maior que ocorre quando a bola j\u00e1 est\u00e1 descendo\u00a0 —\u00a0 t=2,042s\u00a0 —\u00a0 nesse instante a dist\u00e2ncia horizontal da linha de gol ser\u00e1 de X=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.t\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

X=26.0,91.2,042\u00a0 —\u00a0\u00a0X=48,3m<\/b><\/span><\/span><\/span><\/p>\n

12-\u00a0O tempo que a gota de barro permanece no ar \u00e9 o mesmo tempo que a roda demora para efetuar uma volta completa, ou seja, percorrer \u0394S=2\u03c0R com velocidade constante V, que \u00e9 a velocidade de transla\u00e7\u00e3o e de rota\u00e7\u00e3o da roda (n\u00e3o derrapa) e que tamb\u00e9m \u00e9 a velocidade de lan\u00e7amento da gota de barro\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 V=2\u03c0R\/t\u00a0 —\u00a0 t=2\u03c0R\/V\u00a0 —\u00a0 a gota de barro atinge a altura m\u00e1xima h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0na metade desse tempo, quando sua velocidade vertical V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0se anula (V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=V \u2013 g(\u03c0R\/V)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=\u03c0Rg\u00a0 —\u00a0\u00a0V=\u221a(\u03c0Rg)<\/b><\/span><\/span><\/span><\/p>\n

13-\u00a0a) Do gr\u00e1fico\u00a0 —\u00a0 dist\u00e2ncia vertical que percorre\u00a0 at\u00e9 atingir a altura m\u00e1xima\u00a0 —\u00a0 \u0394S=125 \u2013 93,75=31,25cm\u00a0 —\u00a0 \u0394S=0,3125m\u00a0 —\u00a0 na altura m\u00e1xima V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V\u00ad<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2a\u0394S\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10.0,3125\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=2,5m\/s\u00a0 —\u00a0 fun\u00e7\u00e3o hor\u00e1ria vertical\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 quando chega ao solo Y=0\u00a0 —\u00a0 0=0,9375 + 2,5t \u2013 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u00a05t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2,5t \u2013 0,9375=0\u00a0 —\u00a0 \u221a\u0394=5\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

t=(2,5 \u00b15)\/10\u00a0 —\u00a0\u00a0t=0,75s<\/b><\/span><\/span><\/span><\/p>\n

b) Na horizontal\u00a0 —\u00a0 quando X=24m\u00a0 —\u00a0 t=0,75s\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 24=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.0,75\u00a0 —\u00a0V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=32m\/s<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0sem efeito\u00a0 —\u00a0 a for\u00e7a resultante sobre a bola \u00e9 seu peso\u00a0 —\u00a0 P=mg\u00a0 —\u00a0 a=g\u00a0 —\u00a0 com efeito\u00a0 —\u00a0 F=3P (para cima) e P (para baixo)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=3P \u2013 P=2P=2mg\u00a0 —\u00a0 a\u2019=2g\u00a0 —\u00a0 como a acelera\u00e7\u00e3o \u00e9 proporcional \u00e0 velocidade, ela tamb\u00e9m dobrar\u00e1\u00a0 —\u00a0 V\u2019+2.32\u00a0 —\u00a0 V\u2019=64m\/s<\/b><\/span><\/span><\/span><\/p>\n

14-\u00a0a) Y<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 quando t=0,3s\u00a0 —\u00a0 Y=1,2m\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1,2=0 + 0,3V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>= + a.(0,3)<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0,3V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ 0,045a=1,2\u00a0 I<\/b><\/span><\/span><\/span><\/span><\/p>\n

quando t=0,8s\u00a0 —\u00a0 Y=1,2m\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1,2=0 + V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>.0,8 + a(0,8)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0,8V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0,32a = 1,2\u00a0 II\u00a0 —\u00a0 resolvendo o sistema composto por I e II\u00a0 —\u00a0 a=-10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g\u00a0 e V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=5,5m\/s\u00a0 —\u00a0 tempo que demora para atingir a altura m\u00e1xima onde V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 0=5,5 \u2013 10t\u00a0 —\u00a0 t=0,55s\u00a0 —\u00a0 Y<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>= Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>= 0 + 5,5.0,55 \u2013 10(0,55)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0Y<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=1,5125m<\/b><\/span><\/span><\/span><\/p>\n

b) tempo total de movimento t=2.0,55\u00a0 —\u00a0 t=1,1s\u00a0 —\u00a0 na horizontal\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 1,3=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.1,1\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=1,18m\/s<\/b><\/span><\/span><\/span><\/p>\n

c) V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (1,18)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ ((5,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=1,3924 + 37,91\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=6m\/s<\/b><\/span><\/span><\/span><\/p>\n

15-\u00a0a) V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=72km\/h\/3,6=20m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=20.0,35\u00a0 — V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=7m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=20.0,95 \u00a0—\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=19m\/s\u00a0 —\u00a0 tempo que a bola demora para chegar \u00e0 barreira onde X=9,5m com velocidade constante V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=19m\/s\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 t=9,5\/19\u00a0 —\u00a0 t=0,5s\u00a0 —\u00a0 nesse instante a barreira dever\u00e1 ter uma altura vertical de\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=7.0,5 \u2013 5.0,25\u00a0 —\u00a0 Y=3,5 \u2013 1,25\u00a0 —\u00a0\u00a0Y=2,25m<\/b><\/span><\/span><\/span><\/p>\n

b) Tempo que a bola demora para chegar ao gol com velocidade de V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=19m\/s e distante X=19m do ponto de lan\u00e7amento\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0 t=19\/19\u00a0 —\u00a0 t=1s\u00a0 —\u00a0 nesse instante a bola ter\u00e1 uma altura vertical de Y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=7.1 \u2013 5.1\u00a0 —\u00a0 Y=2m (altura da bola ao entrar no gol)\u00a0 —\u00a0 altura da trave=2,4m\u00a0 —\u00a0\u00a0a bola entra no gol 0,4m abaixo da trave.<\/b><\/span><\/span><\/span><\/p>\n

c) Tempo que a bola demora para atingir a altura m\u00e1xima onde V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=7 \u2013 10t\u00a0 —\u00a0 t=0,7s\u00a0 —\u00a0 nesse instante\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t=19.0,7\u00a0 —\u00a0\u00a0X=13,3m\u00a0 —\u00a0 Y=h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>0y<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=7.0,7 \u2013 5.0,49=4,9 \u2013 2,45\u00a0 —\u00a0\u00a0h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>= 2,45m\u00a0 —\u00a0 o tempo que ela<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

demora para retornar ao solo \u00e9 o dobro do tempo que demora para atingir h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t=2.13,3\u00a0 —\u00a0\u00a0t=26,6s\u00a0<\/b><\/span><\/span><\/span><\/p>\n

16-\u00a0a) Colocando o referencial no ponto de lan\u00e7amento e aplicando Torricelli no ponto de altura m\u00e1xima onde v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>e h=1,8m\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2gh\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0-2.10.1,8\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b2=\u221a36\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b2=6\u00a0 —\u00a0 tempo que demora para atingir h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b2 \u2013 10t\u00a0 —\u00a0 0=6 \u2013 10t\u00a0 —\u00a0t=0,6s<\/b><\/span><\/span><\/span><\/p>\n

b) eixo vertical\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b2=6\u00a0 —\u00a0 sen\u03b2=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/6\u00a0 —\u00a0 eixo horizontal\u00a0 —\u00a0 quando t=0,6s\u00a0 —\u00a0 X=3,6m\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0 3,6=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos\u03b2.0,6\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos\u03b2=6\u00a0 —\u00a0 cos\u03b2=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/6\u00a0 —\u00a0 tg\u03b2=sen\u03b2\/cos\u03b2=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/6 x 6\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 tg\u03b2=1\u00a0 —\u00a0\u00a0\u03b2=45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span><\/p>\n

17-\u00a0a) V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos53<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=100.0,60\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=60m\/s\u00a0 —\u00a0\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen53<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=100.0,80\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=80m\/s\u00a0 —\u00a0 quando t=12s\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t=60.12\u00a0 —\u00a0\u00a0X=720m\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=80.12 \u2013 5.(12)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=960 – 720\u00a0 —\u00a0\u00a0Y=240m<\/b><\/span><\/span><\/span><\/p>\n

b) A for\u00e7a resultante \u00e9 o peso do proj\u00e9til, de dire\u00e7\u00e3o vertical e sentido para baixo e de intensidade P=mg=0,1.10\u00a0 —\u00a0 P=1,0N<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

18-\u00a0a) Observe a figura abaixo, onde voc\u00ea deve decompor V em suas componentes vertical V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0e horizontal V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=Vcos\u03b2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=Vcos(\u03b1 + \u03b8)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=Vsen\u03b2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=Vsen(\u03b1 + \u03b8)\u00a0 —\u00a0 equa\u00e7\u00e3o hor\u00e1ria segundo a horizontal X\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0\u00a0X=V.cos (\u03b1 + \u03b8).t\u00a0 —\u00a0 Y=V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0Y=Vsen (\u03b1 + \u03b8).t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

b) Isolando t em X=Vcos(\u03b1 + \u03b8)t\u00a0 —\u00a0 t=X\/Vcos(\u03b1 + \u03b8) que, substitu\u00edda em Y=Vsen(\u03b1 + \u03b8)t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=Vsen(\u03b1 + \u03b8).X\/Vcos(\u03b1 + \u03b8) \u2013 g(X\/Vcos(\u03b1 + \u03b8))<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0Y=tg(\u03b1 + \u03b8) \u2013 g.X<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>cos<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>(\u03b1 + \u03b8)<\/b><\/span><\/span><\/span><\/p>\n

19-\u00a0V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=\u221a2\/2V<\/b><\/span><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=\u221a2\/2V<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 c\u00e1lculo do tempo de subida que ocorre na altura m\u00e1xima quando V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=\u221a2\/2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 t=\u221a2.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2g (tempo de subida)\u00a0 —\u00a0 na horizontal\u00a0 —\u00a0 X=s=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>2t\u00a0 —\u00a0 s=\u221a2\/2.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2(\u221a2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2g)\u00a0 —\u00a0\u00a0s=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/g\u00a0 —\u00a0 na vertical\u00a0 —\u00a0 Y=h==V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=\u221a2\/2.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>(\u221a2.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2g) \u2013 g.(\u221a2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2g)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 h=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2g \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4g\u00a0 —\u00a0\u00a0h=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4g\u00a0 —\u00a0 s\/h=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/g x 4g\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 s\/h=4<\/b><\/span><\/span><\/span><\/p>\n

20- a) V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=100.0,9=90m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=100.0,5=50m\/s\u00a0 —\u00a0 tempo para atingir h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0o que ocorre quando V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=50 \u2013 10t\u00a0 —\u00a0 t=5s\u00a0 —\u00a0 o alcance ocorre em t=2.5\u00a0 —\u00a0 t=10s\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t=90.10\u00a0 —\u00a0 X=900m\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0segundo Simpl\u00edcio\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tg30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=h\/900\u00a0 —\u00a0 \u221a3\/3=h\/900\u00a0 —\u00a0 1,8\/3=h\/900\u00a0 —\u00a0 h=540m<\/b><\/span><\/span><\/span><\/p>\n

c) h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0segundo Salviati\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=50.5 \u2013 5.25\/2=250 – 125\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=125m\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

21- I- V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a mesma (mesmo V<\/b><\/span><\/span><\/span>0\u00a0<\/b><\/span><\/span><\/sub><\/span>e o mesmo \u00e2ngulo)\u00a0 —\u00a0\u00a0 Na h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.g.h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0 = V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2gh<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2g\u00a0 —\u00a0 se g diminui, h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>aumenta\u00a0 —\u00a0 Verdadeira<\/b><\/span><\/span><\/span><\/p>\n

II \u2013 Correta\u00a0 —\u00a0 a velocidade do proj\u00e9til no aponto mais alto da trajet\u00f3ria \u00e9 nula na Terra e na Lua.<\/b><\/span><\/span><\/span><\/p>\n

III \u2013 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a mesma\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 o alcance horizontal X independe de g, assim X \u00e9 o mesmo na Terra e na Lua.<\/b><\/span><\/span><\/span><\/p>\n

IV \u2013 Correta\u00a0 —\u00a0 a velocidade vertical com que ele \u00e9 lan\u00e7ado \u00e9 a mesma, veja I, quem varia \u00e9 g.<\/b><\/span><\/span><\/span><\/p>\n

22-\u00a0Na altura m\u00e1xima\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0 e h=3,2m\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\u2013 2gh\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u2013 2.10 3,2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=8m\/s\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

23-\u00a0I \u2013 Verdadeira\u00a0 —\u00a0 v\u00ea uma composi\u00e7\u00e3o de dois movimentos, um na vertical e outro na horizontal.<\/b><\/span><\/span><\/span><\/p>\n

II \u2013 Falsa\u00a0 —\u00a0 desloca-se em movimento retil\u00edneo uniforme com velocidade horizontal constante.<\/b><\/span><\/span><\/span><\/p>\n

III \u2013 Correta \u2013 na vertical o movimento \u00e9 uniformemente variado com acelera\u00e7\u00e3o a=-g.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

24-\u00a0a) Na h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=1,25m\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2gh<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0-20.1,25\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=5m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=5 \u2013 10t\u00a0 —\u00a0\u00a0t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,5s<\/b><\/span><\/span><\/span><\/p>\n

b) X=V<\/b><\/span><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/span><\/sub><\/span>t=V<\/b><\/span><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/span><\/sub><\/span>2t<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 6=V<\/b><\/span><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/span><\/sub><\/span>.1\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/span><\/sub><\/span>=6,0m\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

c) Trata-se do tempo que ele demora para percorrer na horizontal, com velocidade de V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=6ms a dist\u00e2ncia X=(7,04 \u2013 3,0)=4,04m\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=\u00a0 —\u00a0 4,04=6t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,67s<\/b><\/span><\/span><\/span><\/p>\n

25-\u00a0Bola 1\u00a0 —\u00a0 lan\u00e7amento vertical\u00a0 —\u00a0 tempo para atingir h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0onde V=0\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=30 \u2013 10t\u00a0 —\u00a0 t=3s\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>= 30.3 \u2013 5.9\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>= 45m<\/b><\/span><\/span><\/span><\/p>\n

Bola 2\u00a0 —\u00a0 lan\u00e7amento obl\u00edquo\u00a0 —\u00a0 quando t=3s\u00a0 —\u00a0 h\u2019=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 h\u2019=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=50.1\/2.3 \u2013 10.9\/2=75 \u2013 45\u00a0 —\u00a0\u00a0\u00a0 —\u00a0 h\u2019=30m\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.t=50.\u221a3\/2.3\u00a0 —\u00a0 X=127,5m\u00a0 —\u00a0 a dist\u00e2ncia pedida \u00e9 d, conforme figura abaixo<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(15)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (127,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=\u221a225 + 16.256,25\u00a0 —\u00a0 d=\u221a16.481,25 m\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

26-\u00a0a) V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=5.0,5 \u2013 10t\u00a0 —\u00a0\u00a0t=0,25s<\/b><\/span><\/span><\/span><\/p>\n

b) C\u00e1lculo da altura m\u00e1xima\u00a0 —\u00a0 Y=h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2 + 0,625 – 0,3125\u00a0 —\u00a0h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=2,3125m que \u00e9 menor que a altura da cesta<\/b><\/span><\/span><\/span><\/p>\n

c) na horizontal\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>t\u00a0 —\u00a0 4=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.0,86t —\u00a0 t=4,6\/V<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 na vertical\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 3=2 + 0,5V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.(4,6\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) \u2013 5(4,6\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 1,3=106\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=9,02m\/s<\/b><\/span><\/span><\/span><\/p>\n

27-\u00a0Dados: v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 m\/s; h<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 m;\u00a0q\u00a0= 30\u00b0\u00a0 —\u00a0 as componentes horizontal (v<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>) e vertical (v<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>) da velocidade inicial s\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>\u00a0= v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0cos 30\u00b0 = 10 (0,87) = 8,7 m\/s\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0= vo sem 30\u00b0 = 10 (0,5) = 5 m\/s.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Verificando cada uma das op\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

a) Altura m\u00e1xima atingida em rela\u00e7\u00e3o ao ponto de lan\u00e7amento\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2gh\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2gh\u00a0 —\u00a0 h=V<\/b><\/span><\/span><\/span>oz<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>g=5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/10\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

h=2,5m\u00a0 —\u00a0 em rela\u00e7\u00e3o ao solo\u00a0 —\u00a0 H=2,5 + 10\u00a0 —\u00a0 H=12,5m<\/b><\/span><\/span><\/span><\/p>\n

b) Tempo de subida\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=5 \u2013 10t\u00a0 —\u00a0 t=0,5s<\/b><\/span><\/span><\/span><\/p>\n

c) Com referencial no solo e orientando a trajet\u00f3ria para cima, o tempo para chegar ao solo \u00e9 calculado pela fun\u00e7\u00e3o hor\u00e1ria do espa\u00e7o\u00a0 —\u00a0 h=h<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 h=10 + 5t \u2013 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 quando chega ao solo h=0\u00a0 —\u00a0 0=10 + 5t \u2013 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 t \u2013 5=0\u00a0 —\u00a0 resolvendo a equa\u00e7\u00e3o\u00a0 —\u00a0 t\u00a0@\u00a02,8 s.<\/b><\/span><\/span><\/span><\/p>\n

d) Correta. Ao passar novamente pela mesma altura a pedra possui a mesma energia potencial inicial\u00a0 —\u00a0\u00a0 considerando o sistema<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0conservativo, ent\u00e3o a pedra tem tamb\u00e9m a mesma energia cin\u00e9tica, portanto a mesma velocidade, em m\u00f3dulo, ou seja, se ela \u00e9 lan\u00e7ada com velocidade de 10m\/s, ao retornar passar\u00e1 por esse mesmo ponto com velocidade de -10m\/s.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

28-\u00a0Dados\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 m\/s;\u00a0q\u00a0= 45\u00b0; g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>\u00a0= v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0cos 45\u00b0 = 10.\u221a2\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=5\u221a2m\/s\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0= v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00adsen 45\u00b0 = \u00ad5\u221a2m\/s\u00a0 —\u00a0 no eixo y o movimento \u00e9 uniformemente variado, com a = \u2013g\u00a0 —\u00a0 tempo de subida (t<\/b><\/span><\/span><\/span>sub<\/b><\/span><\/span><\/sub><\/span>), notando que no ponto mais alto v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= v<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 g t\u00a0 —\u00a0 0 = 5\u221a2 \u2013 10 t<\/b><\/span><\/span><\/span>sub<\/b><\/span><\/span><\/sub><\/span>\u00ad\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span><\/span>sub<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u221a2\/2 s\u00a0 —\u00a0 \u00a0tempo\u00a0 de subida \u00e9 igual ao de descida\u00a0 —\u00a0 tempo total (t<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>t=<\/b><\/span><\/span><\/sub><\/span>2t<\/b><\/span><\/span><\/span>sub<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=\u221a2s\u00a0 —\u00a0 no eixo x o movimento \u00e9 uniforme, com velocidade igual a v<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0alcance horizontal (D)\u00a0 —\u00a0 \u00a0D = v<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>\u00a0t<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0= 5.\u221a2.\u221a2\u00a0 —\u00a0 D=10m\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

29-\u00a0Dados: g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>; \u221a2= 1,4;\u00a0q\u00a0= 45\u00b0; v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 20 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

a) Considere desprez\u00edvel a resist\u00eancia do ar e que, ao matar a bola, o p\u00e9 do artilheiro esteja bem pr\u00f3ximo ao ch\u00e3o\u00a0 —\u00a0 ent\u00e3o voc\u00ea pode considerar o ponto de lan\u00e7amento e o ponto de chegada pertencente a um mesmo plano horizontal\u00a0 —\u00a0 no ponto mais alto a componente vertical da velocidade (v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>) \u00e9 nula\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= v<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 g t\u00a0\u00a0\u00de\u00a0 0 = v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u00a0q\u00a0\u2013 g t<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0=20.sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 10t<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=20.\u221a2\/2\/10\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

t<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=\u221a2 s\u00a0 —\u00a0 tempo total =t<\/b><\/span><\/span><\/span>subida<\/b><\/span><\/span><\/sub><\/span>\u00a0+ t<\/b><\/span><\/span><\/span>descida<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>= \u221a2 + \u221a2\u00a0\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=2\u221a2 s\u00a0 —\u00a0 na horizontal o movimento \u00e9 uniforme\u00a0 — velocidade V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0(constante) —\u00a0 v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0 = v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0cos\u00a0q\u00a0= v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0cos 45\u00b0 = 20.\u221a2\/2 m\/s \u00a0—\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=10\u221a2 m\/s\u00a0 —\u00a0 alcance horizontal\u00a0 —\u00a0 x=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>.t=(10\u221a2).(2.\u221a2)\u00a0 —\u00a0\u00a0x=40m<\/b><\/span><\/span><\/span><\/p>\n

b) A velocidade m\u00e9dia (v<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>) do artilheiro pode ser calculada considerando que ele percorreu a dist\u00e2ncia (DS) de 16 m enquanto a bola esteve no ar\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=\u0394S\/\u0394t=16\/2\u221a2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=4\u221a2=4.1,4\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=5,6m\/s=20,16km\/h<\/b><\/span><\/span><\/span><\/p>\n

30-\u00a0Analisando apenas a incorreta, que \u00e9 a 02\u00a0 —\u00a0 a componente horizontal est\u00e1 correta, pois no eixo x o movimento \u00e9 uniforme, por\u00e9m, no eixo y, o movimento \u00e9 uniformemente variado e a equa\u00e7\u00e3o correta \u00e9\u00a0 —\u00a0 y = y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ v<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0t \u2013 \u00a0gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0sen\u03b8\u00a0 —\u00a0 Y=(V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b8)t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 04 + 08 + 16)=29\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

31-\u00a0Para um observador no interior do trem que se desloca em movimento retil\u00edneo e uniforme, o alcance de um objeto lan\u00e7ado horizontalmente s\u00f3 depende da intensidade da velocidade\u00a0 do objeto\u00a0 —\u00a0 assim, caso a bola fosse arremessada em sentido oposto ao do deslocamento do trem, a dist\u00e2ncia entre o ponto de arremesso e o ponto de impacto tamb\u00e9m seria igual a 5 m\u00a0 —\u00a0 n\u00e3o haveria diferen\u00e7a, pois a queda s\u00f3 \u00e9 influenciada por g\u00a0 —\u00a0 logo, seria 5m ao contr\u00e1rio da origem\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

32-\u00a0O tempo de queda \u00e9 calculado exclusivamente pelo movimento vertical (queda livre da altura de 1m com a=g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 h=gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1=10t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 t=\u221a0,2\u00a0 —\u00a0 t=0,447s\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

33-\u00a0O tempo de subida \u00e9 igual ao tempo de descida o que ocorre quando V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b8 \u2013 gt\u00a0 —\u00a0 t=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b8\/g\u00a0 —\u00a0 tempo no ar\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=2t=2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen\u03b8\/g\u00a0 —\u00a0 sendo 2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0e g constantes, o tempo de perman\u00eancia no ar depende apenas do \u00e2ngulo \u03b8 com a horizontal\u00a0 —\u00a0 quanto menor \u03b8, menor ser\u00e1 sen\u03b8 e, consequentemente menor t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

34-<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Decompondo a velocidade inicial\u00a0 —\u00a0 vertical\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\u00a0 —\u00a0 horizontal\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.cos\u03b1\u00a0 —\u00a0 na vertical trata-se de um lan\u00e7amento vertical para cima com velocidade inicial V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1, acelera\u00e7\u00e3o da gravidade (-g) e, na altura m\u00e1xima h=0,30x a velocidade vertical V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-g).h\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0– 2.g.0,30x\u00a0 —\u00a0 x=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1\/6 (I)\u00a0 —\u00a0 tempo de subida\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u2013 gt\u00a0 —\u00a0 0=v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1 \u2013 10t\u00a0 —\u00a0 t=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\/10\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=(tempo de subida + tempo de descida)=2t\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=2.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\/10\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\/5\u00a0 —\u00a0 segundo a horizontal trata-se de movimento retil\u00edneo uniforme com velocidade constante V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.cos\u03b1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=x\/t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos\u03b1=x\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\/5\u00a0 —\u00a0 x=v<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>.cos\u03b1.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\/5\u00a0 —\u00a0 x=v<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.cos\u03b1.sen\u03b1\/5 substituindo (I) em x\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1\/6 = v<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.cos\u03b1.sen\u03b1\/5\u00a0 —\u00a0 sen\u03b1\/6= cos\u03b1\/5\u00a0 —\u00a0 sen\u03b1\/cos\u03b1=6\/5\u00a0 —\u00a0 tg\u03b1=1,2\u00a0 —\u00a0\u00a0 pela tabela \u03b1=50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

35-<\/b><\/span><\/span><\/span><\/p>\n

\u00a002- Colocando os dados fornecidos no sistema de eixos cartesianos\u00a0 —\u00a0 equa\u00e7\u00e3o\u00a0 —\u00a0 y(x) = a(x \u2013 x1).(x \u2013 x2)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

colocando a origem do sistema no ponto de lan\u00e7amento da bola\u00a0 —\u00a0 x1=0 e x2=40)\u00a0 —\u00a0 y(30) = a(x \u2013 0).(x \u2013 40) quando x=30m,\u00a0 y=3m\u00a0 —\u00a0 3 = a.(30 \u2013 0).(30 \u2013 40)\u00a0 —\u00a0 3=a.30.(-10)\u00a0 —\u00a0 a= – (1\/100)\u00a0 —\u00a0 a altura m\u00e1xima ocorre na metade da abscissa x, ou seja, quando x=20 m\u00a0 —\u00a0 y(20)=a.(x \u2013 x1).(x \u2013 x2)\u00a0 —\u00a0 y(20)=- (1\/100). (20 \u2013 0).(20 \u2013 40)\u00a0 —\u00a0 y(20)= – (1\/100).(20).(-20)\u00a0 —\u00a0 y(20)=4 m\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

36-\u00a0Sendo a perda de energia durante a colis\u00e3o desconsiderada neste lan\u00e7amento obliquo voc\u00ea pode considerar\u00a0 todo o movimento horizontal (segundo x) como sendo uniforme de velocidade constante Vox=Vosen\u03b8=10.0,6\u00a0 —\u00a0 Vox=6m\/s\u00a0 —\u00a0 com essa velocidade constante, excluindo o tempo do choque a bola percorre \u2206x=6,6m \u00a0—\u00a0 Vox= \u2206x\/\u2206t\u00a0 —\u00a0 6=6,6\/\u2206t\u00a0 —\u00a0\u00a0 \u2206t=1,1s\u00a0 —\u00a0 o intervalo de tempo total de movimento da bola inclui o tempo do choque (1\/10)s\u00a0 —\u00a0\u00a0 \u2206ttotal=1,1 + 0,1=1,2s\u00a0 —\u00a0\u00a0\u00a0\u2206ttotal=1,2s\u00a0 —\u00a0 c\u00e1lculo da velocidade vertical (Vy) da bola ao chegar ao solo, lan\u00e7ada verticalmente para baixo com Voy=Vo.cos\u03b8=10.0,8\u00a0 —\u00a0 Voy=8m\/s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

Trata-se de um lan\u00e7amento vertical para baixo com velocidade inicial Voy=8m\/s, percorrendo h=1,8m num local onde g=10m\/s2\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 Vy2\u00a0= Voy2\u00a0+ 2.g.h\u00a0 —\u00a0 Vy2=82\u00a0+ 2.10.1,8\u00a0 —\u00a0 Vy=10m\/s\u00a0 —\u00a0 durante a colis\u00e3o vertical, o impulso da for\u00e7a F \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0\u00a0 I=\u2206Q\u00a0 —\u00a0 I=F.\u2206t=F.0,1\u00a0 —\u00a0 \u2206Q=mV \u2013 mVo=0,6.9 \u2013 0,6.(-10)\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\u2206Q=11,4kg.m\/s\u00a0 —\u00a0 I=\u2206Q\u00a0 —\u00a0 0,1F=11,4\u00a0 —\u00a0\u00a0F=114N<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

37-<\/b><\/span><\/span><\/span><\/p>\n

Trata-se de um lan\u00e7amento obl\u00edquo, representado na figura, onde foram colocados os respectivos dados fornecidos no enunciado\u00a0 –<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 c\u00e1lculo de Voy\u00a0 —\u00a0 na altura m\u00e1xima (H=11,25m) a velocidade vertical Vy\u00a0\u00e9 nula (Vy=0)\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Vy2\u00a0= Voy2\u00a0+ 2.(-g).H\u00a0 —\u00a0 02\u00a0= Voy2\u00a0\u2013 2.10.11,25\u00a0 —\u00a0 Voy=\u221a(225)\u00a0 —\u00a0 Voy=15m\/s\u00a0 —\u00a0 tempo de subida\u00a0 —\u00a0 Vy\u00a0= Voy\u00a0\u2013 gt\u00a0 —\u00a0 0=15 \u2013 10t\u00a0 —\u00a0 t=1,5s\u00a0 —\u00a0 para que Prot\u00e1sio consiga alcan\u00e7ar a bola no mesmo n\u00edvel do lan\u00e7amento, ele deve alcan\u00e7a-la no ponto C, onde, desde que partiu, ela demora t=1,5 (metade do percurso horizontal) + 1,5s (outra metade do percurso horizontal)\u00a0 —\u00a0 t=3s\u00a0 —\u00a0 c\u00e1lculo do alcance horizontal x lembrando que,\u00a0 —\u00a0 Vox\u00a0\u00e9 constante e durante todo o movimento horizontal tem intensidade\u00a0 Vox=8m\/s\u00a0 —\u00a0 x=Vox.t=8.3=24m\u00a0 —\u00a0 acelera\u00e7\u00e3o de Prot\u00e1sio\u00a0 —\u00a0 at\u00e9 alcan\u00e7ar a bola ele percorreu \u2206S no mesmo tempo que a bola permaceu no ar, ou seja, t=3s\u00a0 —\u00a0 \u2206S=22,5 \u2013 24<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

=1,5m\u00a0 —\u2206S= Vot + at2\/2\u00a0 —\u00a0 como ele partiu do repouso, Vo=0\u00a0 —\u00a0 1,5=0 + a.32\/2\u00a0 —\u00a0 a=1,5\/4,5\u00a0 —\u00a0 a=1\/3m\/s2\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Voltar para os exerc\u00edcios\u00a0sobre lan\u00e7amento obl\u00edquo<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o dos exerc\u00edcios de vestibulares sobre lan\u00e7amento obl\u00edquo \u00a0 01-\u00a0O tempo que a bola permanece no ar est\u00e1 relacionado\u00a0 com a altura\u00a0 —\u00a0 maior altura, maior tempo de perman\u00eancia no ar\u00a0 —\u00a0\u00a0R- A 02-\u00a0I \u2013 Falsa \u2013 a acelera\u00e7\u00e3o \u00e9 constante e \u00e9 a acelera\u00e7\u00e3o da gravidade\u00a0, sempre com dire\u00e7\u00e3o vertical e sentido para baixo. II \u2013 Correta \u2013 vide<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1085,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1089","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1089","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1089"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1089\/revisions"}],"predecessor-version":[{"id":10797,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1089\/revisions\/10797"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1085"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1089"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}