A ball is dropped from a height h. It bounces | vedclass

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(C) Let the ball be dropped from height $h$. The velocity just before the first impact is $v = \sqrt{2gh}$.After the first impact,the velocity becomes

Vedclass · Accepted Answer

$h\,\left( {\frac{{1 + {e^2}}}{{1 - {e^2}}}} \right)$

Vedclass · Answer

(C) Let the ball be dropped from height $h$. The velocity just before the first impact is $v = \sqrt{2gh}$.After the first impact,the velocity becomes $v_1 = ev = e\sqrt{2gh}$. The height reached is $h_1 = \frac{v_1^2}{2g} = e^2h$.After the second impact,the velocity becomes $v_2 = ev_1 = e^2v$. The height reached is $h_2 = \frac{v_2^2}{2g} = e^4h$.In general,the height reached after the $n^{th}$ impact is $h_n = e^{2n}h$.The total distance $S$ covered by the ball is the sum of the initial fall and the subsequent upward and downward paths:$S = h + 2h_1 + 2h_2 + 2h_3 + ...$$S = h + 2(e^2h + e^4h + e^6h + ...)$$S = h + 2e^2h(1 + e^2 + e^4 + ...)$Using the sum of an infinite geometric series $S_{\infty} = \frac{a}{1-r}$ where $a=1$ and $r=e^2$:$S = h + 2e^2h \left( \frac{1}{1 - e^2} \right)$$S = h \left( 1 + \frac{2e^2}{1 - e^2} \right) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right)$$S = h \left( \frac{1 + e^2}{1 - e^2} \right)$

$A$ ball is dropped from a height $h$. It bounces repeatedly off the ground. What is the total distance covered by the ball before it comes to rest?

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