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Question

(C) ball falling from height $h$ reaches the floor with velocity $v = \sqrt{2gh}$.By the definition of the coefficient of restitution $e$,the velocity

Vedclass · Accepted Answer

$e^4h$

Vedclass · Answer

(C) ball falling from height $h$ reaches the floor with velocity $v = \sqrt{2gh}$.By the definition of the coefficient of restitution $e$,the velocity of separation after the first impact is $v_1 = e \sqrt{2gh}$.The height attained after the first rebound is $h_1 = \frac{v_1^2}{2g} = \frac{(e \sqrt{2gh})^2}{2g} = e^2h$.Similarly,the velocity of separation after the second impact is $v_2 = e v_1 = e^2 \sqrt{2gh}$.The height attained after the second rebound is $h_2 = \frac{v_2^2}{2g} = \frac{(e^2 \sqrt{2gh})^2}{2g} = e^4h$.

$A$ ball of mass $M$ falls from a height $h$ on a floor for which the coefficient of restitution is $e$. The height attained by the ball after two rebounds is

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