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(D) Let the initial height be $h_0$ and the initial velocity just before the first impact be $v_0 = \sqrt{2gh_0}$.The kinetic energy just before the i

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$75$

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(D) Let the initial height be $h_0$ and the initial velocity just before the first impact be $v_0 = \sqrt{2gh_0}$.The kinetic energy just before the impact is $K_i = \frac{1}{2}mv_0^2 = mgh_0$.After the bounce,the velocity becomes $v_1 = ev_0$.The kinetic energy just after the bounce is $K_f = \frac{1}{2}mv_1^2 = \frac{1}{2}m(ev_0)^2 = e^2 K_i$.The loss in kinetic energy is $\Delta K = K_i - K_f = K_i(1 - e^2)$.The percentage loss is $\frac{\Delta K}{K_i} 	imes 100 = (1 - e^2) 	imes 100$.Given $e = 0.5$,the percentage loss $= (1 - (0.5)^2) 	imes 100 = (1 - 0.25) 	imes 100 = 0.75 	imes 100 = 75\%$.

$A$ ball with a coefficient of restitution $e = 0.5$ is dropped from a certain height. What is the percentage loss in kinetic energy after each bounce?

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