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(D) When a ball is dropped from a height $H$ and the coefficient of restitution is $e$,the height reached after the first bounce is $h_1 = e^2 H$,afte

Vedclass · Accepted Answer

$58$

Vedclass · Answer

(D) When a ball is dropped from a height $H$ and the coefficient of restitution is $e$,the height reached after the first bounce is $h_1 = e^2 H$,after the second bounce is $h_2 = e^4 H$,and so on.The total distance $D$ travelled by the ball is given by the sum of the initial fall and the infinite series of bounces (up and down):$D = H + 2h_1 + 2h_2 + 2h_3 + ...$$D = H + 2(e^2 H + e^4 H + e^6 H + ...)$$D = H + 2e^2 H (1 + e^2 + e^4 + ...)$Using the sum of an infinite geometric progression $S = \frac{a}{1-r}$,where $a = 1$ and $r = e^2$:$D = H + 2e^2 H \left( \frac{1}{1 - e^2} ight)$$D = H \left( 1 + \frac{2e^2}{1 - e^2} ight) = H \left( \frac{1 - e^2 + 2e^2}{1 - e^2} ight) = H \left( \frac{1 + e^2}{1 - e^2} ight)$Given $H = 42 \ m$ and $e = 0.4$:$e^2 = (0.4)^2 = 0.16$$D = 42 	imes \left( \frac{1 + 0.16}{1 - 0.16} ight) = 42 	imes \left( \frac{1.16}{0.84} ight)$$D = 42 	imes \frac{116}{84} = 42 	imes \frac{116}{2 	imes 42} = \frac{116}{2} = 58 \ m$.

$A$ ball is allowed to fall freely from a height of $42 \ m$ from the ground. If the coefficient of restitution between the ball and the ground is $0.4$,then the total distance travelled by the ball before it comes to rest is (in $m$)

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