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Question

(A) When a body is dropped from a height $h$,it strikes the floor with velocity $v = \sqrt{2gh}$.After the first impact,it rebounds with velocity $v_1

Vedclass · Accepted Answer

$h\left[\frac{1+e^2}{1-e^2}\right]$

Vedclass · Answer

(A) When a body is dropped from a height $h$,it strikes the floor with velocity $v = \sqrt{2gh}$.After the first impact,it rebounds with velocity $v_1 = ev = e\sqrt{2gh}$.The height reached after the first rebound is $h_1 = \frac{v_1^2}{2g} = e^2h$.After the second impact,it rebounds with velocity $v_2 = ev_1 = e^2v$,reaching a height $h_2 = \frac{v_2^2}{2g} = e^4h$.The total distance $D$ travelled is the initial fall plus twice the sum of all subsequent rebound heights:$D = h + 2h_1 + 2h_2 + 2h_3 + \dots$$D = h + 2(e^2h + e^4h + e^6h + \dots)$$D = h + 2e^2h(1 + e^2 + e^4 + \dots)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=e^2$:$D = h + 2e^2h \left( \frac{1}{1-e^2} \right)$$D = h \left[ 1 + \frac{2e^2}{1-e^2} \right] = h \left[ \frac{1-e^2+2e^2}{1-e^2} \right] = h \left[ \frac{1+e^2}{1-e^2} \right]$.

$A$ body falls freely from a height $h$ on a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution,the total distance travelled before it comes to rest is

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