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(C) When a ball is dropped from a height $h$,its velocity just before the first impact is $v_1 = \sqrt{2gh}$.After the first impact with the ground,th

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$he^4$

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(C) When a ball is dropped from a height $h$,its velocity just before the first impact is $v_1 = \sqrt{2gh}$.After the first impact with the ground,the velocity becomes $v_1' = e v_1 = e \sqrt{2gh}$,where $e$ is the coefficient of restitution.The height reached after the first impact is $h_1 = \frac{(v_1')^2}{2g} = \frac{e^2 (2gh)}{2g} = he^2$.After the second impact,the velocity becomes $v_2' = e v_1' = e(e \sqrt{2gh}) = e^2 \sqrt{2gh}$.The height reached after the second impact is $h_2 = \frac{(v_2')^2}{2g} = \frac{(e^2 \sqrt{2gh})^2}{2g} = \frac{e^4 (2gh)}{2g} = he^4$.Therefore,the ball reaches a height of $he^4$ after the second impact.

$A$ ball is dropped from a height $h$. After it strikes the ground twice,what height will it reach?

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