A ball is dropped from a height h. It bounces | vedclass

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Question

(A) Let the initial velocity of the ball just before the first impact be $v = \sqrt{2gh}$. The time taken to fall from height $h$ is $t = \sqrt{2h/g}$

Vedclass · Accepted Answer

$e^{2n}h, e^nt$

Vedclass · Answer

(A) Let the initial velocity of the ball just before the first impact be $v = \sqrt{2gh}$. The time taken to fall from height $h$ is $t = \sqrt{2h/g}$.After the first collision with the ground,the velocity of the ball becomes $v_1 = ev$,where $e$ is the coefficient of restitution.The height attained after the first bounce is $h_1 = v_1^2 / (2g) = (ev)^2 / (2g) = e^2 (v^2 / 2g) = e^2 h$.The time taken to reach the maximum height after the first bounce is $t_1 = v_1 / g = ev / g = et$.After the second collision,the velocity becomes $v_2 = ev_1 = e^2v$.The height attained after the second bounce is $h_2 = v_2^2 / (2g) = (e^2v)^2 / (2g) = e^4 (v^2 / 2g) = e^4 h$.The time taken to reach the maximum height after the second bounce is $t_2 = v_2 / g = e^2v / g = e^2t$.Following this pattern,after $n$ bounces,the height attained is $h_n = e^{2n}h$.The time taken to reach the maximum height after the $n$-th bounce is $t_n = e^nt$.

$A$ ball is dropped from a height $h$. It bounces repeatedly. Find the height attained by the ball after $n$ bounces and the time taken by the ball to complete $n$ bounces.

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