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(A) When a ball is dropped from a height $h$ and undergoes an inelastic collision with the ground,the coefficient of restitution $e$ is defined as the

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$h e^6$

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(A) When a ball is dropped from a height $h$ and undergoes an inelastic collision with the ground,the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.After the first collision,the velocity of the ball becomes $v_1 = ev_0$,where $v_0 = \sqrt{2gh}$ is the velocity just before the first impact.The height reached after the first collision is $h_1 = \frac{v_1^2}{2g} = \frac{(ev_0)^2}{2g} = e^2 h$.Similarly,after the second collision,the height reached is $h_2 = e^2 h_1 = e^2 (e^2 h) = e^4 h$.Following this pattern,after the $n^{th}$ collision,the height reached is $h_n = h e^{2n}$.For the third collision $(n = 3)$,the height reached is $h_3 = h e^{2(3)} = h e^6$.

$A$ tennis ball is released from height $h$ above ground level. If the ball makes an inelastic collision with the ground,to what height will it rise after the third collision?

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