A body falls freely onto a hard horizontal su | vedclass

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(A) Let the initial height from which the body falls be $H_0$. When a body falls from height $H_0$,its velocity just before the first impact is $v_0 =

Vedclass · Accepted Answer

$256: 625$

Vedclass · Answer

(A) Let the initial height from which the body falls be $H_0$. When a body falls from height $H_0$,its velocity just before the first impact is $v_0 = \sqrt{2gH_0}$. After the first impact,the velocity becomes $v_1 = e v_0$,and the height reached is $H_1 = e^2 H_0$. After the second impact,the velocity becomes $v_2 = e v_1 = e^2 v_0$,and the height reached is $H_2 = e^4 H_0$. Given $e = 0.8 = 4/5$. The ratio of the height after the second impact to the initial height is $H_2 / H_0 = e^4$. Calculating $e^4 = (0.8)^4 = (4/5)^4 = 256 / 625$. Thus,the ratio is $256: 625$.

$A$ body falls freely onto a hard horizontal surface. If the coefficient of restitution between the surface and the body is $e = 0.8$,then the ratio of the maximum height to which the body rises after the second impact to the initial height of the body is:

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