A ball falls freely from a height h on a rigi | vedclass

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Question

(B) When the ball is dropped from height $h$,it strikes the plane with velocity $v = \sqrt{2gh}$.After the collision,the ball rebounds with a velocity

Vedclass · Accepted Answer

$h(1+2e^2)$

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(B) When the ball is dropped from height $h$,it strikes the plane with velocity $v = \sqrt{2gh}$.After the collision,the ball rebounds with a velocity $u = ev = e\sqrt{2gh}$.The height reached by the ball after the first collision is $h_1 = \frac{u^2}{2g} = \frac{e^2(2gh)}{2g} = e^2h$.The total distance travelled by the ball before hitting the plane for the second time is the sum of the initial downward distance and the distance covered during the first rebound (up and down).Total distance $= h + 2h_1 = h + 2(e^2h) = h(1 + 2e^2)$.

$A$ ball falls freely from a height $h$ on a rigid horizontal plane. If the coefficient of restitution is $e$,then the total distance travelled by the ball before hitting the plane the second time is

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