{"id":1047,"date":"2015-07-15T03:56:46","date_gmt":"2015-07-15T03:56:46","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1047"},"modified":"2024-08-23T12:36:00","modified_gmt":"2024-08-23T12:36:00","slug":"resolucoes-dos-exercicios-sobre-lancamento-vertical-para-cima","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/lancamento-vertical-para-cima\/resolucoes-dos-exercicios-sobre-lancamento-vertical-para-cima\/","title":{"rendered":"Lan\u00e7amento vertical para cima"},"content":{"rendered":"

Resolu\u00e7\u00f5es dos exerc\u00edcios sobre lan\u00e7amento vertical para cima<\/b><\/span><\/span><\/span><\/p>\n

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01- a) Chegar\u00e3o juntos, pois n\u00e3o existe mat\u00e9ria (ar), retardando-os.<\/b><\/span><\/span><\/p>\n

b) A pessoa que est\u00e1 com o p\u00e1ra-quedas fechado chegar\u00e1 primeiro, pois o ar retardar\u00e1 a pessoa com o p\u00e1ra-quedas aberto, devido \u00e0 maior \u00e1rea de contato com o mesmo.<\/b><\/span><\/span><\/p>\n

02-\u00a0<\/b><\/span><\/span>R- D\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 Na superf\u00edcie da Lua o meio \u00e9 o v\u00e1cuo<\/b><\/span><\/span><\/p>\n

03-\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>\u00a0 —\u00a0 independente da massa, se n\u00e3o existe ar, os corpos caem com a mesma acelera\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

04- Como em cada instante elas tem a mesma acelera\u00e7\u00e3o e consequentemente a mesma velocidade, elas percorrem a mesma dist\u00e2ncia no mesmo tempo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

05- Em todos os pontos da trajet\u00f3ria a acelera\u00e7\u00e3o \u00e9 sempre constante e de valor g, inclusive no ponto mais alto da trajet\u00f3ria e onde a velocidade \u00e9 nula, pois ele p\u00e1ra para transformar seu movimento de progressivo em retr\u00f3grado\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

06- Na altura m\u00e1xima V=0 e \u0394S=h<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>=50cm=0,5m\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u0394S\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0– 2.10.0,5\u00a0 — \u00a0V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u221a10=3,16ms\u00a0 —<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

07- Um movimento vertical independe da massa\u00a0 —\u00a0 altura m\u00e1xima=3,2m e nela V=0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0– 2.g.h<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.10.3,2\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u221a64\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=8ms\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

08- a)\u00a0 Pelo diagrama a part\u00edcula precisa de 1 s para atingir a altura m\u00e1xima (v = 0), que \u00e9 o ponto onde a reta intercepta o eixo t. Ser\u00e1 necess\u00e1rio mais 1 s para pedra retornar ao ponto de partida, pois o tempo de subida \u00e9 igual ao ponto de descida\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- 2s<\/b><\/span><\/span><\/p>\n

b) V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10m\/s\u00a0 —\u00a0\u00a0 a altura m\u00e1xima em rela\u00e7\u00e3o ao solo, que ocorre quando t=1s, vale\u00a0 —\u00a0 S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t \u2013 g.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=10.1 \u2013 5.1\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= 5m (altura m\u00e1xima em rela\u00e7\u00e3o ao ponto de lan\u00e7amento)\u00a0<\/b><\/span><\/span><\/p>\n

A pedra demora 3s para chegar ao solo\u00a0 —\u00a0 S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t \u2013 g.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=10.3 \u2013 5.9\/2\u00a0 —\u00a0 S=30 \u2013 45\u00a0 —\u00a0 S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= -15m (posi\u00e7\u00e3o em que chega ao solo)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- 15m<\/b><\/span><\/span>\u00a0 —\u00a0 Veja figura abaixo<\/b><\/span><\/span><\/p>\n

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\"\"<\/p>\n

c) Veja figura acima\u00a0 —\u00a0 H=15 + 5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- H=20m<\/b><\/span><\/span><\/p>\n

09- Atinge o solo com velocidade de -30m\/s, pois j\u00e1 est\u00e1 em movimento retrogrado\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 g.t\u00a0 —\u00a0 -30=10 \u2013 10.t\u00a0 —\u00a0 t=4s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

10- Colocando a origem no ponto de lan\u00e7amento, a pedra sobe, atinge a altura m\u00e1xima e quando chega ao solo ocupa a posi\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

S = – 60m\u00a0 — instante em que ela chega ao solo\u00a0 —\u00a0 S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t \u2013 g.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 -60=20t \u2013 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0-4t -12=0\u00a0 —\u00a0 t=6s\u00a0 (instante em que<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

ela chega ao solo)\u00a0 —\u00a0 velocidade dela quando chega ao solo (t=6s)\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 gt\u00a0 —\u00a0 V=20 \u2013 10.6\u00a0 —\u00a0 V=-40m\/s\u00a0 —\u00a0 em m\u00f3dulo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=40m\/s<\/b><\/span><\/span><\/p>\n

11- C\u00e1lculo dos instantes em que ele passa pela altura de 14m, ou seja, S=14m\u00a0 —\u00a0 S=10 + 5t \u2013 t<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 14=10 +5t \u2013 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 5t + 4=0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 t=1s (subida e t\u2019=4s (descida)\u00a0 —\u00a0 ele estar\u00e1 acima de 14m (luminosidade \u00fatil) entre os instantes 1s e 4s\u00a0 —\u00a0 \u0394t=4 \u2013 1=3s\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

—\u00a0 t=1s (subida e t\u2019=4s (descida)\u00a0 —\u00a0 ele estar\u00e1 acima de 14m (luminosidade \u00fatil) entre os instantes 1s e 4s\u00a0 —\u00a0 \u0394t=4 \u2013 1=3s\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

12- \u0394S=25cm=0,25m\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=?\u00a0 —\u00a0 a altura atingida independe da massa do foguetinho\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.g.\u0394S\u00a0 —\u00a0 na altura m\u00e1xima V=0\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.10.0,25\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u221a5=2,236m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

13- V=0\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=?\u00a0 —\u00a0 g=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394S=3,2m\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.g.\u0394S\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 20.3,2\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=8m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

14- No instante em que a l\u00e2mpada se desprende do teto do elevador ele est\u00e1 sendo lan\u00e7ada para cima com velocidade de V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=5m\/s e o tempo de subida e descida at\u00e9 chegar ao piso, na posi\u00e7\u00e3o \u2013x, foi t=0,5s\u00a0 —\u00a0 (veja figura abaixo)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 -x=5.0,5 \u2013 5.(0,5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 x= – 1,25m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>altura do elvador = 1,25m<\/b><\/span><\/span><\/p>\n

15- C\u00e1lculo do tempo de subida\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 gt<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0=20 \u2013 10t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=2s\u00a0 —\u00a0 rela\u00e7\u00e3o entre o\u00a0 tempo de subida t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=2s e o de descida t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\/t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=2\/3\u00a0 —\u00a0 2\/t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=2\/3\u00a0 —\u00a0 t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=3s\u00a0 —\u00a0 tempo total do movimento (subida + descida)\u00a0 —\u00a0 t=2 + 3\u00a0\u00a0 —\u00a0 t=5s\u00a0 —\u00a0 S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

S=20.5 \u2013 5.25\u00a0 —\u00a0 S= -25m\u00a0 —<\/b><\/span><\/span>\u00a0 h=25m<\/b><\/span><\/span><\/p>\n

16- a) Cada bola fica no ar t= 3.0,4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=1,2s<\/b><\/span><\/span>
\nb) Como cada bola fica no ar 1,2s e o tempo de subida \u00e9 igual ao tempo de descida, ela demora t=0,6s para atingir a altura m\u00e1xima,\u00a0 onde V=0\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 gt\u00a0 —\u00a0 0=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 10.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=6m\/s<\/b><\/span><\/span><\/p>\n

c) S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=6.0,6 \u2013 5.(0,6)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0S=1,8m<\/b><\/span><\/span>
\n17- a) O bal\u00e3o sobe com velocidade constante V e, aos 5s atinge a altura h, onde a pedra \u00e9 abandonada com\u00a0\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 V=h\/5\u00a0 —\u00a0 essa velocidade \u00e9 a velocidade com que a pedra \u00e9 abandonada, ou seja, \u00e9 a velocidade inicial com que a pedra sobe\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=h\/5
\n\"\"<\/b><\/span><\/span><\/p>\n

Observe que o movimento de sobe e desce da pedra demora 2s e que ela chega ao solo quando t=2s e na posi\u00e7\u00e3o S=-h\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

S-V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 -h=h\/5.2 \u2013 4,9.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=98\/7\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0h=14m<\/b><\/span><\/span><\/p>\n

c) C\u00e1lculo da velocidade de subida do bal\u00e3o que vale V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=h\/5=14\/5=2,8m\/s\u00a0 —\u00a0 V=2,8m\/s\u00a0 —\u00a0 o bal\u00e3o subiu durante 7s com velocidade constante de 2,8m\/s e percorreu\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 2,8=H\/7\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>H=19,6m<\/b><\/span><\/span>
\n18- Entre 0 e 1s (primeiro segundo), entre 1s e 2s (segundo segundo) e entre 2s e 3s (terceiro segundo), ou seja,\u00a0 no decorrer do terceiro segundo quer dizer entre\u00a0 t=2s e t=3s\u00a0 —\u00a0 t=2s\u00a0 —\u00a0 S<\/b><\/span><\/span>2s<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.2 \u2013 5.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 20\u00a0 —\u00a0 S<\/b><\/span><\/span>2s<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 20\u00a0 —\u00a0 t=3s\u00a0 — S<\/b><\/span><\/span>3s<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t \u2013 gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.3 \u2013 5.3<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>3s<\/b><\/span><\/span><\/sub>=3V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 45\u00a0 —\u00a0 \u0394S=15m (do enunciado)\u00a0 —\u00a0 \u0394S=S<\/b><\/span><\/span>3s<\/b><\/span><\/span><\/sub>\u00a0\u2013 S<\/b><\/span><\/span>2s<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 15=3V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 45 \u2013 (2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 20)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=40m\/s<\/b><\/span><\/span>
\n19- a) Bolinha P que saiu antes com V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=15m\/s\u00a0 —\u00a0 tempo que demora para atingir a altura m\u00e1xima, ou seja, quando V=0\u00a0 —\u00a0 V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 gt\u00a0 —\u00a0 0=15 \u2013 10t\u00a0 —\u00a0 t=1,5s (tempo de subida e tempo de descida)\u00a0 —\u00a0 se ela parte com velocidade de 15ms ela retorna ao ponto de partida com velocidade de -15m\/s (veja gr\u00e1fico abaixo)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A bolinha Q repete os mesmos movimentos de Q, mas depois de 0,5s (veja gr\u00e1fico acima)\u00a0<\/b><\/span><\/span><\/p>\n

b) Como P partiu 0,5s antes\u00a0 —\u00a0 t<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=0,5\u00a0 —\u00a0 t<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0+ 0,5\u00a0 —\u00a0 \u00a0S<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=15t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0-5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=15t<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0-5t<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=15(t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0+ 0,5) – 5(t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0+ 0,5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=15t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0+ 7,5 \u2013 5(t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>.0,5 + 0,25)=15t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0+ 7,5 \u2013 5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0– 5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0\u2013 1,25\u00a0 —\u00a0 S<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=10t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0\u2013 5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 6,25\u00a0 —\u00a0 no encontro S<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

15t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0\u2013 5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=-5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 10t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0+ 6,25\u00a0 —\u00a0 t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=6,25\/5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=1,25s (instante do encontro ap\u00f3s a partida de Q) e t<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=1,75s (instante do encontro ap\u00f3s a partida de P)<\/b><\/span><\/span>\u00a0 —\u00a0 substituindo t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=1,25s em S<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=15t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0\u2013 5t<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=15.1,25 \u2013 5.(1,25)<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>S<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=11m<\/b><\/span><\/span><\/p>\n

20- Na altura m\u00e1xima V=0\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0– 2.g.h\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0-2.10.0,45\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

21- O corpo que foi lan\u00e7ado para cima com velocidade V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, quando retorna ao ponto de partida tem velocidade \u2013V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>. Assim, como os dois s\u00e3o lan\u00e7ados para baixo com velocidades de mesmo m\u00f3dulo, eles chegar\u00e3o ao solo com a mesma velocidade em m\u00f3dulo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

22-\u00a0<\/b><\/span><\/span>R- C\u00a0<\/b><\/span><\/span>\u00a0(veja teoria)<\/b><\/span><\/span><\/p>\n

23-\u00a0<\/b><\/span><\/span>R- D\u00a0<\/b><\/span><\/span>\u00a0\u00a0(veja teoria)<\/b><\/span><\/span><\/p>\n

24- I \u2013 Falsa \u2013 os instantes devem ser os mesmos para percorrerem a mesma dist\u00e2ncia<\/b><\/span><\/span><\/p>\n

II- Falsa \u2013 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=m.g.h \u2013 a energia potencial gravitacional depende da massa<\/b><\/span><\/span><\/p>\n

III \u2013 Correta \u2013 essa acelera\u00e7\u00e3o \u00e9 a da gravidade<\/b><\/span><\/span><\/p>\n

IV \u2013 Correta \u2013 ambos os fen\u00f4menos est\u00e3o relacionados com a lei fundamental da atra\u00e7\u00e3o gravitacional de Newton \u2013 F=G.M.m\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

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

25-\u00a0<\/b><\/span><\/span>O lan\u00e7amento vertical, livre de resist\u00eancia do ar, \u00e9 um movimento uniformemente variado. A velocidade varia com o tempo de acordo com a fun\u00e7\u00e3o: v = v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\"\"g t\u00a0 —\u00a0 portanto, o gr\u00e1fico \u00e9 uma reta, sendo o m\u00f3dulo da velocidade decrescente na subida, crescente na descida e nulo no ponto mais alto.\u00a0<\/b><\/span><\/span><\/p>\n

R- A\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

26-\u00a0<\/b><\/span><\/span>Equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0\u00a0 v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2a\u0394S\u00a0 —\u00a0 no ponto mais alto\u00a0 —\u00a0 V=0 e \u0394S=h\u00a0 —\u00a0 a=-g\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2gh\u00a0 —\u00a0 h=v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

27-\u00a0<\/b><\/span><\/span>a) Aplicando a equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2a\u0394S\u00a0 —\u00a0 no ponto mais alto\u00a0 —\u00a0 V=0 e \u0394S=h\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=8<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.1,6.h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>h=20,0m \u00a0<\/b><\/span><\/span>\u00a0—\u00a0 c\u00e1lculo do tempo de subida\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 gt\u00a0 —\u00a0 0=8 \u2013 1,6t\u00a0 —\u00a0 t=5,0s\u00a0 —\u00a0 tempo total=tempo de subida + tempo de descida\u00a0 —\u00a0 t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=5 + 5\u00a0 —\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=10,0s.<\/b><\/span><\/span><\/p>\n

b) Na Terra, a pena chega depois porque o efeito da resist\u00eancia do ar sobre ela \u00e9 mais significativo que sobre o martelo\u00a0 —\u00a0 por\u00e9m a Lua \u00e9 praticamente desprovida de atmosfera, e n\u00e3o havendo for\u00e7as resistivas significativas, o martelo e a pena caem com a mesma acelera\u00e7\u00e3o, atingindo o solo lunar ao mesmo tempo, como demonstrou David Randolph Scott em seu experimento.\u00a0<\/b><\/span><\/span><\/p>\n

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

Colocando a origem dos espa\u00e7os no ponto de lan\u00e7amento (h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0) e orientando a trajet\u00f3ria para cima\u00a0 —\u00a0 lan\u00e7ado para baixo\u00a0 —\u00a0 h<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=0 -10t – gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2= -10.1 + 10.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>= – 15m\u00a0 —\u00a0 lan\u00e7ado para cima\u00a0 —\u00a0 h<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=0 + 10.1 \u2013 10.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=5m\u00a0 —\u00a0 dist\u00e2ncia que os<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

separa\u00a0 —\u00a0 d=5 \u2013 (-15)\u00a0 —\u00a0 d=20m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

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

–<\/b><\/span><\/span>\u00a0Colocando a origem no ponto de lan\u00e7amento, a pedra sobe, atinge a altura m\u00e1xima e quando chega ao solo ocupa a posi\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

S = – 90m\u00a0 — instante em que ela chega ao solo\u00a0 —\u00a0 h=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t \u2013 g.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 -90=15t \u2013 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0– 3t -18=0\u00a0 —\u00a0 t=6s\u00a0 (instante em que<\/b><\/span><\/span><\/p>\n

ela chega ao solo)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

30-\u00a0<\/b><\/span><\/span>\u00a0a) Pelo enunciado o VLS acelera verticalmente at\u00e9 atingir a altura de 100m e, a partir desse instante (t=0), ele se move por alguns segundos com velocidade constante de 20m\/s at\u00e9 que um dos conjuntos de instrumentos se desprende, o que ocorre, pelo gr\u00e1fico, quando t=2s\u00a0 —\u00a0 observando o gr\u00e1fico voc\u00ea pode concluir que quando t=0 ele atinge a velocidade de 20m\/s\u00a0 —\u00a0 portanto a ordenada y vale y=V=20m\/s\u00a0 —\u00a0 entre 0 e 2s ele subiu com velocidade constante de 20ms\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V=\u2206S\/\u2206t\u00a0 —\u00a0 20=\u2206S\/2\u00a0 —\u00a0 \u2206S=h\u2019=40m\u00a0 —\u00a0 como, desde o solo at\u00e9 atingir a velocidade de 20m\/s ele j\u00e1 havia subido h=100m, a altura total alcan\u00e7ada foi de htotal=100 + 4==140m\u00a0 —\u00a0\u00a0h=140m.<\/b><\/span><\/span><\/p>\n

b) No instante em que o conjunto de instrumentos se desprende ele possui a mesma velocidade para cima que o VLS e que \u00e9 de 20m\/s\u00a0 —\u00a0 assim, o movimento do conjunto de instrumentos \u00e9 como um lan\u00e7amento vertical para cima com velocidade inicial Vo=20m\/s\u00a0 —\u00a0 agora o conjunto est\u00e1 subindo e sua velocidade est\u00e1 variando de Vo=20m\/s\u00a0 (quando t=2s) paraV=0 (quando t=4s), veja gr\u00e1fico\u00a0 —\u00a0 a=g=\u2206V\/\u2206t=(V \u2013 Vo)\/(t \u2013 to)=(0 \u2013 20)\/(4 \u2013 2)\u00a0 —\u00a0\u00a0g= – 10m\/s2\u00a0\u00a0—\u00a0 a partir do instante t=4s, em que o conjunto de instrumentos se desprende, ele possui a mesma velocidade que a do VLS que \u00e9 de 20m\/s\u00a0 —\u00a0 ent\u00e3o, trata-se de um lan\u00e7amento vertical para cima com Vo=20m\/s, a=g= -10m\/s2\u00a0 —\u00a0 colocando a origem no ponto de lan\u00e7amento, quando o conjunto de instrumentos atinge o solo ele est\u00e1 na posi\u00e7\u00e3o S= – 140m\u00a0 —\u00a0 S= So\u00a0+ Vo.t + g.t2\/2\u00a0 —\u00a0 -140 =0 + 20t -10.t2\/2\u00a0 —\u00a0 t2\u00a0\u2013 4t \u2013 28=0\u00a0 —\u00a0 \u2206=16 + 112=128\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

t= (4 \u00b1 11,31)\/2\u00a0 —\u00a0 t\u22487,6s\u00a0 —\u00a0 o tempo pedido, vale t=2 + 7,6\u00a0 —\u00a0\u00a0t\u22489,6s<\/b><\/span><\/span><\/p>\n

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

Caracter\u00edsticas do lan\u00e7amento vertical com a origem no ponto de lan\u00e7amento e a trajet\u00f3ria orientada para cima\u00a0 —\u00a0 na subida, o movimento \u00e9 progressivo, pois o deslocamento ocorre no sentido crescente da trajet\u00f3ria, e retardado, pois o m\u00f3dulo da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

velocidade est\u00e1 diminuindo\u00a0 —na descida, o movimento \u00e9 retrogrado, pois o deslocamento ocorre no sentido decrescente da trajet\u00f3ria, e acelerado, pois o m\u00f3dulo da velocidade est\u00e1 aumentando\u00a0 —\u00a0 no ponto mais alto da trajet\u00f3ria, a velocidade do corpo se anula (V=0), pois \u00e9 o ponto em que o corpo inverte o sentido de seu movimento e nesse ponto a altura atingida pelo corpo \u00e9 m\u00e1xima\u00a0 —\u00a0 o tempo de subida \u00e9 igual ao tempo de descida\u00a0 —\u00a0 a velocidade (V\u00ad\u00ad<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) de lan\u00e7amento na origem \u00e9 igual \u00e0 mesma velocidade de chegada \u00e0 origem, mas de sinal contr\u00e1rio (-Vo)\u00a0 —em qualquer ponto da trajet\u00f3ria o corpo tem duas velocidades de mesmo m\u00f3dulo, uma positiva na subida e uma negativa na descida.<\/b><\/span><\/span><\/p>\n

Analisando cada assertativa:<\/b><\/span><\/span><\/p>\n

I. Correta\u00a0 —\u00a0 a acelera\u00e7\u00e3o \u00e9 sempre a mesma em qualquer instante e \u00e9 negativa, pois a concavidade da par\u00e1bola \u00e9 para baixo.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 a velocidade do corpo tem intensidade m\u00e1xima no instante de lan\u00e7amento (t=0) e de volta ao mesmo ponto de lan\u00e7amento (t=4s).<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 \u00e9 o ponto onde a altura \u00e9 m\u00e1xima (h=20m) e a velocidade \u00e9 nula\u00a0 —\u00a0 nesse instante (t=2s) ele p\u00e1ra e inverte o sentido de seu movimento.<\/b><\/span><\/span><\/p>\n

IV. Correta\u00a0 —\u00a0 ele sobe retardando, p\u00e1ra e a partir de t=2s, desce acelerando.<\/b><\/span><\/span><\/p>\n

R- D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00f5es dos exerc\u00edcios sobre lan\u00e7amento vertical para cima   01- a) Chegar\u00e3o juntos, pois n\u00e3o existe mat\u00e9ria (ar), retardando-os. b) A pessoa que est\u00e1 com o p\u00e1ra-quedas fechado chegar\u00e1 primeiro, pois o ar retardar\u00e1 a pessoa com o p\u00e1ra-quedas aberto, devido \u00e0 maior \u00e1rea de contato com o mesmo. 02-\u00a0R- D\u00a0\u00a0—\u00a0 Na superf\u00edcie da Lua o meio \u00e9 o v\u00e1cuo<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1043,"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-1047","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1047","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=1047"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1047\/revisions"}],"predecessor-version":[{"id":10800,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1047\/revisions\/10800"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1043"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1047"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}