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Exercício de Lançamentos
Exercício:
Um objeto é lançado verticalmente para cima no local onde o modulo da aceleração da gravidade local é 10m/s². Sabendo-se que, quando atinge um terçço de sua altura maxima, tem uma velocidade de 8m/s, pode-se afirmar que a altura maxima atingida pelo objeto, em metros, é igual a:
a) 4,8
b) 3,9
c) 3,2
d) 2,6
e) 1,7
A resposta é letra a : 4,8
Informações Adicionais:
Pessoal, passei horas tentando fazer essa questaum dar 4,8, so consigu ter o resultado igual a 3,2.. Pessoal, me ajudem nessa questaum, por favor…
Atenciosamente, Mateus Ramos…
Eu tentei e tentei, eu consegui atraves de uns conceito, agora non sei se esta certo….
Geralmente fiz o seguinte:Utilizei a formula da função horaria da velocidade pra descobrir o tempo e depois utilizei a formula da queda livre pra calcular do ponto mais alto ate o um terço da altura:
v= v(inicial) +gt
0=8 +10t
t= 0,8
h=gt²/2
h=10.(0,8)² /2
h= 3,2 ( esse valor da altura é apenas, o acima de um terço da altura)
Agora calculei, a do 1/3 da altura, que no caso foi lançamnto vertical…
h= h(inicial) + v(inicial).t – gt²/2 ( o g foi negativo pq no lançamento vertical a gravidade é negativa pois a velociade diminui quando sobe)
h= 0 + 8.(0,8) – 10. (0.8)² /2
h= 6,4 – 3,2
h= 3,2 ( so que ele quer um terço dessa altura)
enton fikou…
1/3.h
1/3×3,2
o valor deu 1,6…
Agora juntando o as 2 alturas, deu:
3,2 + 1,6 = 4,8 a altura total..
Se estiver errado, por favor me ajudem a tirar a duvida…
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