A ball falls from a height h and rebounds aft | vedclass

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(C) When a ball falls from a height $h$,the height reached after the $n^{th}$ rebound is given by $h_n = h e^{2n}$,where $e$ is the coefficient of res

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) When a ball falls from a height $h$,the height reached after the $n^{th}$ rebound is given by $h_n = h e^{2n}$,where $e$ is the coefficient of restitution.The total distance $H$ covered by the ball before it comes to rest is the sum of the initial fall and the subsequent upward and downward paths for each rebound:$H = h + 2h_1 + 2h_2 + 2h_3 + \dots$$H = h + 2(h e^2) + 2(h e^4) + 2(h e^6) + \dots$$H = h + 2h(e^2 + e^4 + e^6 + \dots)$The term in the bracket is an infinite geometric progression with first term $a = e^2$ and common ratio $r = e^2$. The sum is $S = \frac{a}{1-r} = \frac{e^2}{1-e^2}$.Substituting this back into the expression for $H$:$H = h + 2h \left( \frac{e^2}{1-e^2} \right)$$H = h \left( 1 + \frac{2e^2}{1-e^2} \right)$$H = h \left( \frac{1 - e^2 + 2e^2}{1-e^2} \right)$$H = h \left( \frac{1 + e^2}{1-e^2} \right)$

$A$ ball falls from a height $h$ and rebounds after striking the floor. The coefficient of restitution is $e$. The total distance covered before it comes to rest is

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