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(D) The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.For a ball dropped from heigh

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$he^{2n}$

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(D) The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.For a ball dropped from height $h$,the velocity just before the first impact is $u = \sqrt{2gh}$.After the first impact,the velocity of rebound is $v_1 = e \cdot u = e\sqrt{2gh}$.The height reached after the first bounce is $h_1 = \frac{v_1^2}{2g} = \frac{(e\sqrt{2gh})^2}{2g} = e^2h$.Similarly,after the second bounce,the height reached is $h_2 = e^2 h_1 = e^2(e^2h) = e^4h$.Following this pattern,after the $n^{th}$ bounce,the height reached is $h_n = e^{2n}h$.

$A$ ball is dropped from a height $h$ onto a ground level. If the coefficient of restitution is $e$,what is the height reached by the ball after the $n^{th}$ bounce?

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