{"id":1060,"date":"2015-07-19T03:44:02","date_gmt":"2015-07-19T03:44:02","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1060"},"modified":"2024-08-23T12:48:40","modified_gmt":"2024-08-23T12:48:40","slug":"resolucao-dos-exercicios-sobre-queda-livre-vertical","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/queda-livre-vertical\/resolucao-dos-exercicios-sobre-queda-livre-vertical\/","title":{"rendered":"Queda livre vertical – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o dos exerc\u00edcios sobre queda livre vertical<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01<\/b><\/span><\/span>– Como a resist\u00eancia do ar \u00e9 desprezada eles caem com a mesma acelera\u00e7\u00e3o que \u00e9 a da gravidade e consequentemente suas velocidades, em cada instante, ser\u00e3o sempre as mesma, independente das massas e consequentemente dos pesos\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0V=28m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=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<\/b><\/span><\/span><\/sup>\u00a0+ 2.g.\u0394S\u00a0 —\u00a0 28<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.9,8.\u0394S\u00a0 —\u00a0 \u0394S=784\/19,6\u00a0 —\u00a0 H=40m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0S= gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=5.16\u00a0 —\u00a0 S=80m\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0Segundo toco B no instante 1s caiu\u00a0 —\u00a0 S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=5.1=5m\u00a0\u00a0 —\u00a0 como o primeiro toco partiu 1s antes, quando t=1s para B, ser\u00e1 2s para A\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= 5t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=5.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=20m\u00a0 —\u00a0 \u0394S=20 \u2013 5=15m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.g.h\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.500\u00a0 —\u00a0 V=100ms\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0Chegam ao solo com a mesma velocidade\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0a) Como o bal\u00e3o se desloca horizontalmente, sua velocidade vertical inicial \u00e9 nula\u00a0 —\u00a0 S=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 80=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=4s<\/b><\/span><\/span><\/p>\n

b) Corresponde \u00e0 dist\u00e2ncia horizontal que o bal\u00e3o percorre em 4s com velocidade constante de 6m\/s\u00a0 —\u00a0 V=d\/t\u00a0 —\u00a0 6=d\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=24m<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0I- Correta \u2013 5m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0significa 5(ms)\/s, ou seja, varia\u00e7\u00e3o de velocidade de 5ms em cada 1s<\/b><\/span><\/span><\/p>\n

II \u2013 Falsa –\u00a0 a cada segundo o deslocamento vai aumentando, pois a velocidade \u00e9 cada vez maior.<\/b><\/span><\/span><\/p>\n

III \u2013 Falsa \u2013 segundo o enunciado a acelera\u00e7\u00e3o \u00e9 constante e vale 5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

IV \u2013 Verdadeira \u2013 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ g.t= 0 + 5.8\u00a0 —\u00a0 V=40m\/s<\/b><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span>\u00a0Tempo que a pedra demora para cair h=7,2m\u00a0 —\u00a0 S=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 7,2=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=1,2s\u00a0 —\u00a0 nesse tempo o carro com velocidade constante de 120km\/h=33,4m\/s o carro se deslocou\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 33,4= \u0394S\/1,2\u00a0 —\u00a0 \u0394S=40,08m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0a)Corpo que \u00e9 abandonado da altura H e demora T para chegar ao ch\u00e3o\u00a0 —\u00a0 H=10T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 corpo que \u00e9 abandonado da altura H\/4 e que demora (T \u2013 3)s para chegar ao ch\u00e3o\u00a0 —\u00a0 H\/4=10.(T \u2013 3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 dividindo membro a membro\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/(T \u2013 3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0=4\u00a0 —\u00a0 T\/T \u2013 3 =2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=6s<\/b><\/span><\/span><\/p>\n

b) H=10T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=10.36\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0H=180m<\/b><\/span><\/span><\/p>\n

11-\u00a0<\/b><\/span><\/span>A velocidade inicial de descida (V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) do objeto \u00e9 a mesma que do helic\u00f3ptero\u00a0 —\u00a0 S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + g.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 100=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.4 + 10.16\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=20\/4\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=5m\/sX3,6=18km\/h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0a) Nesse caso, pela equa\u00e7\u00e3o observava-se que a origem da trajet\u00f3ria est\u00e1 no solo e que ela \u00e9 orientada para cima\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0comparando S=S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 com S=180 \u2013 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=180m ou, substituindo t=0 na equa\u00e7\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- 180m<\/b><\/span><\/span><\/p>\n

a) A altura do pr\u00e9dio \u00e9 dada pela fun\u00e7\u00e3o em t = 0, ou seja, 180 metros.<\/b><\/span><\/span><\/p>\n

b) A pedra atinge o solo quando S=0\u00a0 —\u00a0 0= 180 – 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, isto \u00e9, 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 180 ==> t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 180\/5 = 36, de onde vem que<\/b><\/span><\/span>\u00a0t = 6 s.<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0S=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 11=4t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t=\u221a5,5=2,3s\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

Colocando a origem no ponto de lan\u00e7amento e orientando a trajet\u00f3ria para baixo, 1uando ele se encontra a 80m do solo, sua posi\u00e7\u00e3o \u00e9 S=130 \u2013 80=50m<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- t=2s<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0a) A acelera\u00e7\u00e3o de queda da plataforma \u00e9 a acelera\u00e7\u00e3o da gravidade local fornecida pelo exerc\u00edcio,<\/b><\/span><\/span>\u00a0g=10m\/<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>
\nb) Velocidade da plataforma quando se encontra 30m acima do solo, ou seja, ap\u00f3s ter percorrido \u0394S=75 \u2013 30=45m\u00a0 —\u00a0 Torricelli\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.g.\u0394S\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.45\u00a0 —\u00a0<\/b><\/span><\/span>V=30m\/s<\/b><\/span><\/span><\/p>\n

c)<\/b><\/span><\/span>\u00a0Imobilizada (freando \u2013 a<0)\u00a0 —\u00a0 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<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-a).h\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=30<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2a.30\u00a0 —\u00a0 a=900\/60\u00a0 —\u00a0 a=15m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>
\n18-<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.g.h=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.5\u00a0 —\u00a0 V=10m\/sX3,6=36km\/h\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span>\u00a0Propor\u00e7\u00f5es de Galileu para queda livre com V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 e desprezando-se a resist\u00eancia do ar:<\/b><\/span><\/span><\/p>\n

Para intervalos de tempos iguais e consecutivos, um corpo em queda livre percorre dist\u00e2ncias cada vez maiores, na propor\u00e7\u00e3o dos<\/b><\/span><\/span><\/p>\n

\u00edmpares consecutivos: no primeiro segundo, o m\u00f3vel cai uma dist\u00e2ncia d; no segundo seguinte, percorre 3d; no terceiro segundo,<\/b><\/span><\/span><\/p>\n

5d, e assim por diante.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) falsa \u2013 velocidade \u00e9 diferente de acelera\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

b) falsa \u2013 estava a 0,50m do solo<\/b><\/span><\/span><\/p>\n

c) falsa \u2013 estava a 0,80m do solo<\/b><\/span><\/span><\/p>\n

d) Verdadeira \u2013 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 quanto maior a velocidade, maior ser\u00e1 a energia cin\u00e9tica\u00a0 —\u00a0 o terceiro pingo tem maior velocidade que o quarto pingo, que, nesse instante est\u00e1 abandonando a torneira.<\/b><\/span><\/span><\/p>\n

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

20-<\/b><\/span><\/span>\u00a0a) 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.g.h=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.1\u00a0 —\u00a0 V=\u221a20\u00a0 —\u00a0 V=4,47m\/s<\/b><\/span><\/span><\/p>\n

b) Como o tempo de queda \u00e9 o mesmo, o intervalo de tempo entre as batidas de duas gotas consecutivas \u00e9 o mesmo intervalo de tempo compreendido entre as sa\u00eddas de duas gotas consecutivas da torneira. Se saem 3 gotas em cada 60s, tem-se que, entre a 1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0e a 2<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0(20s); entre a 2<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0e a 3<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0(20s) e entre a 3<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0e a 4<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0(20s) e assim, a partir do instante em que uma bate no solo, depois de 20s chega a outra ao solo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- 20s<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0Objeto M, solto do repouso V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0\u00a0 —\u00a0 S<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=H=gt<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 H=5t<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 objeto N arremessado para baixo com velocidade inicial V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=80m\/s\u00a0 —\u00a0 S=V<\/b><\/span><\/span>o.<\/b><\/span><\/span><\/sub>t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ gt<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S=80t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>+ 5t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como M partiu 4s antes\u00a0 —\u00a0 t<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=4\u00a0 —\u00a0 t<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 4\u00a0 — H=15.(t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 4)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 H=5(t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 8t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 16)\u00a0 —\u00a0 H=5t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2\u00a0<\/b><\/span><\/span><\/sup>+ 40t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ \u00a080\u00a0 —\u00a0 se encontram no solo\u00a0 —\u00a0 H = S\u00a0 —\u00a0 5t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2\u00a0<\/b><\/span><\/span><\/sup>+ 40t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 80 = 80t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 5t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 40t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 80 = 80t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=2s\u00a0 —\u00a0 substituindo t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=2s em S=H=80t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ 5t<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=80.2 + 5.4=160 + 20\u00a0 —<\/b><\/span><\/span>\u00a0 H=180m\u00a0 — R- D<\/b><\/span><\/span><\/p>\n

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

C\u00e1lculo da velocidade V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0ap\u00f3s a 1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0colis\u00e3o\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.g.h<\/b><\/span><\/span>O<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u221a2gh<\/b><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sub>I\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 velocidade logo ap\u00f3s a 1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0colis\u00e3o\u00a0 —\u00a0 f=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>1\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=f.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(velocidade de subida que \u00e9 igual em m\u00f3dulo \u00e0 velocidade V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0com que ela retorna)\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=f.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

F=V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=f.f.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00ad\u00a0 —\u00a0 V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=f<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 analogamente ap\u00f3s a\u00a0 3<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0colis\u00e3o\u00a0 —\u00a0 V<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=f<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 assim, ap\u00f3s a 3<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0colis\u00e3o ela \u00e9<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

lan\u00e7ada para cima com velocidade V<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=f<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0atingindo uma altura h, quando V=0\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= V<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2gh\u00a0 —\u00a0 0 = (f<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.\u221a2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.g.h\u00a0 —\u00a0\u00a0\u00a0 f<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.2gh<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 2gh\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>h=f<\/b><\/span><\/span>6.<\/b><\/span><\/span><\/sup>h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/p>\n

23-<\/b><\/span><\/span>\u00a0I \u2013 Falsa\u00a0 —\u00a0 a velocidade aumenta de 10ms em cada 1s<\/b><\/span><\/span><\/p>\n

II \u2013 Falsa\u00a0 —\u00a0 h=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h=5.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=5m<\/b><\/span><\/span><\/p>\n

III \u2013 Verdadeira\u00a0 —\u00a0 V=gt=10.2\u00a0 —\u00a0 V=20m\/s<\/b><\/span><\/span><\/p>\n

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

24-<\/b><\/span><\/span>\u00a0Tempo de queda do saco\u00a0 —\u00a0 h=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 tempo o avi\u00e3o deve se deslocar \u0394S com velocidade constante de V=144km\/h\/3,6=40m\/s\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 40= \u0394S\/10\u00a0 —\u00a0 \u0394S=400m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span>\u00a0A velocidade \u00e9 a mesma independente da massa\u00a0 —\u00a0 P<\/b><\/span><\/span>melancia<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>melancia<\/b><\/span><\/span><\/sub>.g=5g\u00a0 —\u00a0 P<\/b><\/span><\/span>ma\u00e7\u00e3<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>ma\u00e7\u00e3<\/b><\/span><\/span><\/sub>.g=0,05.g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span>\u00a0H=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 20=5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t=2s\u00a0 —\u00a0 V=gt=10.2=20m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

27-\u00a0<\/b><\/span><\/span>Observe a figura abaixo onde L \u00e9 a dist\u00e2ncia horizontal entre a mancha e o dubl\u00ea no instante do salto\u00a0 —\u00a0 c\u00e1lculo do tempo de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0queda do dubl\u00ea\u00a0 —\u00a0 h=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 5=10t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1\u00a0 —\u00a0 t=1s\u00a0 —\u00a0 a velocidade ideal V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=\u0394S\/\u0394t=(L + 3)\/1\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=L + 3\u00a0 —\u00a0 velocidade m\u00ednima\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t=L\/1\u00a0 —\u00a0 V<\/b><\/span><\/span>min<\/b><\/span><\/span><\/sub>=L\u00a0 —\u00a0 velocidade m\u00e1xima\u00a0 —\u00a0 V<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t=(L + 6)\/1\u00a0 —\u00a0 V<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>=L + 6\u00a0 —\u00a0 diferen\u00e7as\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0– V<\/b><\/span><\/span>min<\/b><\/span><\/span><\/sub>= (L + 3) \u2013 L\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= V<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>\u00a0 – V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=(L + 6) \u2013 (L + 3)\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3m\/s<\/b><\/span><\/span><\/p>\n

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

28-\u00a0<\/b><\/span><\/span>Dados: h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 20 m; g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>; v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 0.<\/b><\/span><\/span><\/p>\n

A Figura abaixo\u00a0 ilustra a situa\u00e7\u00e3o descrita no enunciado\u00a0 —\u00a0 desprezando a resist\u00eancia do ar, trata-se de uma queda livre, que \u00e9<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

um movimento uniformemente variado\u00a0 — \u00a0equa\u00e7\u00e3o da altura em fun\u00e7\u00e3o do tempo\u00a0 —\u00a0 h=h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>2\u00a0 —\u00a0 h=20 \u2013 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

gr\u00e1fico dessa fun\u00e7\u00e3o \u00e9 um arco de par\u00e1bola com concavidade para baixo\u00a0 —\u00a0 quando h=0\u00a0 —\u00a0 0=20 \u2013 5t2\u00a0 —\u00a0 t=\u221a4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=2s<\/b><\/span><\/span><\/p>\n

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

29-\u00a0<\/b><\/span><\/span>Dados: g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0; t = 6 s\u00a0 —\u00a0 para a queda livre\u00a0 —\u00a0 h=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=10.6<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h=180m\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

30-\u00a0<\/b><\/span><\/span>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>\u00a0+ 2.g.h\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2gh\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>cTerra<\/b><\/span><\/span><\/sub>=m.(2g<\/b><\/span><\/span>Terra<\/b><\/span><\/span><\/sub>.h)\u00a0 —\u00a0 E<\/b><\/span><\/span>cLua<\/b><\/span><\/span><\/sub>=m.(2g<\/b><\/span><\/span>Lua<\/b><\/span><\/span><\/sub>.h)\u00a0 —\u00a0 como a gravidade na superf\u00edcie da Terra \u00e9 maior que a gravidade na superf\u00edcie da Lua, a energia cin\u00e9tica na superf\u00edcie da Terra \u00e9 maior que a energia cin\u00e9tica na superf\u00edcie da Lua\u00a0 —\u00a0 em rela\u00e7\u00e3o \u00e0 queda livre\u00a0 —\u00a0 h=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t=\u221a2h\/g\u00a0 —\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 como g<\/b><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><g<\/b><\/span><\/span>Terra<\/b><\/span><\/span><\/sub>, o tempo de queda na Lua \u00e9 maior do que na Terra\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

C\u00e1lculo da dist\u00e2ncia total (\u2206S) percorrida\u00a0 —\u00a0 \u2206S=5 + 15 + 25 + 35=80m\u00a0 —\u2206S=80m\u00a0 —\u00a0 trata-se de uma queda livre com S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 e V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 (abandonado)\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>+ gt=0 + 10.4\u00a0 —\u00a0 V=40m\/s.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

A. Falsa \u00a0—\u00a0 a energia mec\u00e2nica diminui se houver atrito com o ar ou permanece a mesma se n\u00e3o houver atrito com o ar.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

B. Falsa\u00a0 —\u00a0 a velocidade da r\u00e9gua aumenta a medida que ela desce.<\/b><\/span><\/span><\/p>\n

C. Falsa\u00a0 —\u00a0 a acelera\u00e7\u00e3o da r\u00e9gua \u00e9 a acelera\u00e7\u00e3o da gravidade que permanece constante (mesmo local).<\/b><\/span><\/span><\/p>\n

D. Correta\u00a0 —\u00a0 a for\u00e7a que provoca a queda da r\u00e9gua \u00e9 a for\u00e7a peso, que a faz descer com acelera\u00e7\u00e3o constante que \u00e9 a da gravidade g, pois P=m.g.<\/b><\/span><\/span><\/p>\n

E. Falsa\u00a0 —\u00a0 veja (B).<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Resolu\u00e7\u00e3o dos exerc\u00edcios sobre queda livre vertical   01– Como a resist\u00eancia do ar \u00e9 desprezada eles caem com a mesma acelera\u00e7\u00e3o que \u00e9 a da gravidade e consequentemente suas velocidades, em cada instante, ser\u00e3o sempre as mesma, independente das massas e consequentemente dos pesos\u00a0 —\u00a0 R- B 02-\u00a0V=28m\/s\u00a0 —\u00a0 Vo=0\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V2= Vo2\u00a0+ 2.g.\u0394S\u00a0 —\u00a0 282=02\u00a0+ 2.9,8.\u0394S\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1056,"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-1060","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1060","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=1060"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1060\/revisions"}],"predecessor-version":[{"id":10808,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1060\/revisions\/10808"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1056"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1060"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}