{"id":1100,"date":"2015-08-02T14:21:00","date_gmt":"2015-08-02T14:21:00","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1100"},"modified":"2024-08-23T12:29:12","modified_gmt":"2024-08-23T12:29:12","slug":"resolucao-comentada-dos-exercicios-sobre-lancamento-horizontal","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/lancamento-horizontal\/resolucao-comentada-dos-exercicios-sobre-lancamento-horizontal\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios LAN\u00c7AMENTO HORIZONTAL"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios sobre \u00a0<\/b><\/span><\/span><\/span>LAN\u00c7AMENTO HORIZONTAL<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span>\u00a0\u2013 veja teoria<\/b><\/span><\/span><\/p>\n

02-\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span>\u00a0\u2013 em todo lan\u00e7amento horizontal, a componente V<\/b><\/span><\/span>0x<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=constante.<\/b><\/span><\/span><\/p>\n

03- O tempo que o destr\u00f3ier disp\u00f5e \u00e9 o tempo que a bomba demora para cair e percorrer a dist\u00e2ncia vertical de 8km\u00a0 —\u00a0 Y=8.000m\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 8.000=10t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t=\u221a1.600\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=40s<\/b><\/span><\/span><\/p>\n

04- Tempo que a pedra demora para atingir a superf\u00edcie da \u00e1gua, ou seja, percorrer verticalmente Y=1m\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

1=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=\u221a0,2s\u00a0 —\u00a0 em t=\u221a0,2=0,45s ela percorre na horizontal X=3m\u00a0 —\u00a0 X=V<\/b><\/span><\/span>o.t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 3=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.0,45\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=6,7m\/s<\/b><\/span><\/span><\/p>\n

05- Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 0,8=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=0,4s\u00a0 —\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t\u00a0 —\u00a0 0,8=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.0,4\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2,0m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

06- tempo de queda\u00a0 —\u00a0 Y=gt2\/2\u00a0 —\u00a0 5=5t2\u00a0 —\u00a0 t=1s (\u00e9 sempre o mesmo, depende apenas da altura vertical)\u00a0 —\u00a0 velocidade m\u00ednima (limite inferior de R)\u00a0 —\u00a0 X=V.t\u00a0 —\u00a0 1=V.1\u00a0 —\u00a0 V=1ms\u00a0 —\u00a0 limite posterior (velocidade m\u00e1xima)\u00a0 —\u00a0 X=V\u2019.t\u00a0 —\u00a0 4=V\u2019.1\u00a0 —\u00a0 V\u2019=4ms\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

07- O tempo t de queda aumenta, pois a altura aumenta\u00a0 — como X1=1m e X2=4m permanecem constantes, V1 e V2 devem diminuir para manter a igualdade X=V.t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

08- Velocidade vertical com que ela chega ao solo\u00a0 —\u00a0 V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=2.g.h=2.10.0,6\u00a0 —\u00a0 V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=12\u00a0 —\u00a0 velocidade horizontal com que ela chega ao solo\u00a0 —\u00a0 \u00e9 a mesma de lan\u00e7amento\u00a0 —\u00a0 V<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=2m\/s\u00a0 —\u00a0 velocidade resultante V com que ela chega ao solo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A intensidade da velocidade resultante com que ela chega ao solo vale V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ V<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=12 + 4\u00a0 —\u00a0 V=\u221a16\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=4m\/s<\/b><\/span><\/span>\u00a0 —\u00a0 mas, como o choque \u00e9 perfeitamente el\u00e1stico, ela retorna com velocidade resultante de mesma intensidade, ou seja, V\u2019=V=4m\/s\u00a0 —\u00a0 no tri\u00e2ngulo hachurado da figura acima\u00a0 —\u00a0 sen\u03b2=V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\/V\u2019=\u221a12\/4=2\u221a3\/4=\u221a3\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03b2=60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

09- A ordem de chegada ao solo, independente das massas , depende da altura de queda\u00a0 —\u00a0 menor altura, menos tempo de queda\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

10- a) Lan\u00e7amento horizontal do sistema (bloco + proj\u00e9til) que independe da massa\u00a0 —\u00a0 tempo que demora para chegar ao solo na vertical percorrendo a altura Y=1,25m\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 1,25=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=0,5s\u00a0 —\u00a0 nesse tempo ele percorre na horizontal X=5m com velocidade V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t\u00a0 —\u00a0 5=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10m\/s<\/b><\/span><\/span><\/p>\n

b) choque inel\u00e1stico\u00a0 —\u00a0 movem-se juntos ap\u00f3s o mesmo\u00a0 —\u00a0 princ\u00edpio da conserva\u00e7\u00e3o da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

quantidade de movimento\u00a0 — quantidade de movimento\u00a0 –Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=0,02.V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0+ 0,480.0\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=0,02.V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=(m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>).V=(0,02 + 0,480).10\u00a0 —\u00a0 Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=5kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>depois\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 0.02V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=5\u00a0 —\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=250m\/s<\/b><\/span><\/span><\/p>\n

11- N\u00e3o, pois enquanto o alvo permanece im\u00f3vel, o proj\u00e9til, durante todo seu movimento estar\u00e1 caindo, sujeito \u00e0 acelera\u00e7\u00e3o da gravidade e chegar\u00e1 ao alvo numa posi\u00e7\u00e3o abaixo daquela em que ele mirou.<\/b><\/span><\/span><\/p>\n

12- Como o alvo permanece im\u00f3vel, o proj\u00e9til, durante todo seu movimento estar\u00e1 caindo, sujeito \u00e0 acelera\u00e7\u00e3o da gravidade e chegar\u00e1 ao alvo numa posi\u00e7\u00e3o abaixo daquela em que ele mirou\u00a0 —\u00a0 tempo que o proj\u00e9til demora para percorrer a dist\u00e2ncia horizontal de X=30m com V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=600m\/s (constante)\u00a0 —\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t\u00a0 —\u00a0 30=600t\u00a0 —\u00a0 t=0,05s\u00a0 —\u00a0 nesse tempo ele desce uma altura Y de\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>2=5.(0,05)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Y=5.(0,0025)=0,0125m=1,25cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

13- Dist\u00e2ncia vertical (altura do pr\u00e9dio) percorrida em 2s\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=5.4=20m\u00a0 —\u00a0 n\u00famero de andares\u00a0 —\u00a0 n=altura do pr\u00e9dio\/altura de cada andar=20\/2,5\u00a0 —\u00a0 n=8 andares\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

14- tg45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=D\/H\u00a0 —\u00a0 1=D\/H\u00a0 —\u00a0 D=H\u00a0 — na vertical\u00a0 —\u00a0\u00a0 H=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 H=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 na horizontal\u00a0 —\u00a0 X=Vt\u00a0 —\u00a0 D=H=10t\u00a0 —<\/b><\/span><\/span><\/p>\n

10t=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=2s\u00a0 —\u00a0 D=H=10.2=20m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

15- V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=144km\/h=144\/3,6\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=40m\/s\u00a0 — tempo que o saco demora para chegar ao solo na dire\u00e7\u00e3o vertical\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 500=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=10s\u00a0 —\u00a0 nesse instante o alvo deve estar a uma dist\u00e2ncia horizontal de X=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t=40.10\u00a0 —\u00a0 X=400m\u00a0 —<\/b><\/span><\/span>–\u00a0 R- D<\/b><\/span><\/span><\/p>\n

16- tempo que demora para percorrer Y=0,80m na vertical\u00a0 —\u00a0 Y=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 0,80=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=0,4s\u00a0 —\u00a0 nesse tempo ele percorre na horizontal X=2,8m com velocidade V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t\u00a0 —\u00a0 2,8=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.0,4\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=7,0m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span>R- E\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

18- V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=360\/3,6=100m\/s\u00a0 —\u00a0 t=0\u00a0 —\u00a0 X<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 t=1s\u00a0 —\u00a0 X<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t=100.1=100m\u00a0 —\u00a0 t=2s\u00a0 —\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t=100.2=200m\u00a0 —\u00a0 a dist\u00e2ncia entre os pontos de impacto ser\u00e1 de<\/b><\/span><\/span>\u00a0100 m.<\/b><\/span><\/span><\/p>\n

19-\u00a0<\/b><\/span><\/span>Por in\u00e9rcia as tr\u00eas caixas continuaram em movimento com a mesma velocidade horizontal do avi\u00e3o de 360 km\/h\u00a0 —\u00a0 desta forma os impactos no solo ocorrer\u00e3o sobre a mesma linha reta, separadas pela dist\u00e2ncia percorrida pelo avi\u00e3o durante aquele 1 s entre os lan\u00e7amentos das caixas\u00a0 —\u00a0 \u00a0velocidade de 360 km\/h corresponde a 100 m\/s e desta forma a dist\u00e2ncia entre os pontos de impacto ser\u00e1 de 100 m.<\/b><\/span><\/span><\/p>\n

20-\u00a0<\/b><\/span><\/span>(01) Correta\u00a0 —\u00a0 a trajet\u00f3ria somente \u00e9 parab\u00f3lica quando a resist\u00eancia do ar \u00e9 nula.<\/b><\/span><\/span><\/p>\n

(02) Errada\u00a0 —\u00a0 a componente horizontal da velocidade s\u00f3 se mant\u00e9m constante se a resist\u00eancia do ar \u00e9 nula.<\/b><\/span><\/span><\/p>\n

(04) Errada\u00a0 —\u00a0 a for\u00e7a de arrasto tem sentido oposto ao da velocidade em cada ponto, sempre tangente \u00e0 trajet\u00f3ria.<\/b><\/span><\/span><\/p>\n

(08) Errada\u00a0 —\u00a0 mesma justificativa de (01) e (02).<\/b><\/span><\/span><\/p>\n

(16) Correta\u00a0 —\u00a0 \u00a0componente horizontal da for\u00e7a de resist\u00eancia do ar faz diminuir a componente horizontal da velocidade, que tende para zero, enquanto a componente vertical da resist\u00eancia do ar tende a equilibrar o peso. Assim o movimento tende a se tornar retil\u00edneo e uniforme, na dire\u00e7\u00e3o vertical.\u00a0<\/b><\/span><\/span><\/p>\n

R- (01+ 16) = 17<\/b><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/span><\/span>Com a mola comprimida de X a energia potencial el\u00e1stica armazenada na mesma vale\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=kX<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —ao deixar o contato com a mola toda a energia potencial el\u00e1stica \u00e9 transformada em energia cin\u00e9tica de valor E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 considerando o sistema conservativo (sem atrito) essas energias se igualam\u00a0 —\u00a0 KX<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=KX<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/m\u00a0 —\u00a0 utilizando agora a conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 no topo do plano\u00a0 — \u00a0E<\/b><\/span><\/span>MA<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + mgh\u00a0 — no solo chega com velocidade V\u2019\u00a0 —\u00a0 E<\/b><\/span><\/span>MB<\/b><\/span><\/span><\/sub>\u00a0= mV\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + mg.0\u00a0 —\u00a0 E<\/b><\/span><\/span>MA<\/b><\/span><\/span><\/sub>\u00a0= E<\/b><\/span><\/span>MB<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + mgh=mV<\/b><\/span><\/span>\u20192<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0\u00a0 V\u2019=\u221a(2gh + KX<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/m)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span>Quando a menina abandona a bola, a menina continua em movimento retil\u00edneo e a bola executa um lan\u00e7amento horizontal\u00a0 —\u00a0 as velocidades horizontais da bola e da menina s\u00e3o as mesmas e valem V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=10,8\/3,6=3m\/s\u00a0 —\u00a0 ent\u00e3o, a dist\u00e2ncia horizontal percorrida pela menina e pela bola \u00e9 a mesma\u00a0 —\u00a0 S<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=V.\u0394t=3.0,5\u00a0 —\u00a0 S<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1,5m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Utilize as informa\u00e7\u00f5es a seguir para responder \u00e0s quest\u00f5es de n\u00fameros 23 e 24<\/b><\/span><\/span><\/p>\n

Tr\u00eas bolas \u2212 X, Y e Z \u2212 s\u00e3o lan\u00e7adas da borda de uma mesa, com velocidades iniciais paralelas ao solo e mesma dire\u00e7\u00e3o e sentido.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A tabela acima mostra as magnitudes das massas e das velocidades iniciais das bolas.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Trata-se de um lan\u00e7amento horizontal que segundo o eixo y \u00e9 uma queda livre com V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=0 e sob acelera\u00e7\u00e3o da gravidade g\u00a0 —\u00a0 h=V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>t + gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=0 + gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 observe que, como h e g s\u00e3o os mesmos, o tempo de queda tamb\u00e9m ser\u00e1 o mesmo\u00a0 —<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Em todo lan\u00e7amento horizontal a velocidade horizontal \u00e9 constante, igual \u00e0 velocidade de lan\u00e7amento, independente da massa do corpo\u00a0 —\u00a0 assim, a esfera que \u00e9 lan\u00e7ada com maior velocidade consegue o maior alcance\u00a0 —<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

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

Como o exerc\u00edcio pede a altura voc\u00ea deve utilizar apenas o movimento parcial na dire\u00e7\u00e3o vertical (eixo Y)<\/b><\/span><\/span><\/p>\n

A proje\u00e7\u00e3o da acelera\u00e7\u00e3o da gravidade (\"\")\u00a0\u00e9 o pr\u00f3prio\u00a0\"\".\u00a0 Assim, na dire\u00e7\u00e3o vertical tem-se um movimento uniformemente variado (MUV), ou seja, um lan\u00e7amento vertical para cima, com acelera\u00e7\u00e3o igual \u00e0 acelera\u00e7\u00e3o da gravidade\u00a0\"\"\u00a0e velocidade inicial de lan\u00e7amento\u00a0 Voy, tal que Voy=Vo.sen\u03b1, onde \u03b1 \u00e9 o \u00e2ngulo de lan\u00e7amento..\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.sen\u03b1 – componente vertical de V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Y – altura num instante qualquer t<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 Y<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 altura inicial<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 g – acelera\u00e7\u00e3o da gravidade<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 orientando o eixo Y para cima tem-se um lan\u00e7amento vertical de equa\u00e7\u00f5es\u00a0 —\u00a0 S=Y\u00a0 —\u00a0 S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Y<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.sen\u03b1\u00a0 —\u00a0 a=-g<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 equa\u00e7\u00e3o da posi\u00e7\u00e3o (dos espa\u00e7os) ou da altura Y<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 S=S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>.t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 Y=0 + V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>.t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 — \u00a0<\/b><\/span><\/span>Y= V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>.t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2, com V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=Vo.sen\u03b1.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.sen\u03b8=30.0,5\u00a0 —\u00a0 V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=15m\/s\u00a0 —\u00a0 Y=15.1 \u2013 10.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 Y=10m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\u00a0Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios sobre \u00a0LAN\u00c7AMENTO HORIZONTAL   01-\u00a0R- E\u00a0\u2013 veja teoria 02-\u00a0R- C\u00a0\u2013 em todo lan\u00e7amento horizontal, a componente V0x=Vo=Vx=constante. 03- O tempo que o destr\u00f3ier disp\u00f5e \u00e9 o tempo que a bomba demora para cair e percorrer a dist\u00e2ncia vertical de 8km\u00a0 —\u00a0 Y=8.000m\u00a0 —\u00a0 Y=gt2\/2\u00a0 —\u00a0 8.000=10t2\/2\u00a0 —\u00a0 t=\u221a1.600\u00a0 —\u00a0\u00a0t=40s 04- Tempo que a pedra demora para<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1096,"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-1100","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1100","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=1100"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1100\/revisions"}],"predecessor-version":[{"id":10795,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1100\/revisions\/10795"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1096"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}