a ball is thrown upward with an initial vertical velocity of v0 to a maximum hei

Question

a ball is thrown upward with an initial vertical velocity of v0 to a maximum height of hmax and falls back toward earth. on the way down, it is caught at the same height at which it was thrown upward. if the time the ball remains in the air is t, what is the maximum height the ball attained while in the air?

1) hmax= 4gt^2
2) hmax= (1/4g)t^2
3) hmax= 8gt^2
4) hmax= (1/8)gt^2
5) hmax= gt^2
6) hmax= 2gt^2
7) hmax= (1/2)gt^2

Explanation / Answer

given initial velocity =0 so v=u+gt => v=gt now from conservation of energy 0.5*m*v^2 = m*g*h => 0.5*(gt)^2 = g*h => h=0.5*g*t^2 answer if u still have any doubt post it as comment have a nice day !!!

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