A ball is dropped from a height of 5 m. The | vedclass

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(C) The time taken for the first drop is $t_0 = \sqrt{\frac{2h}{g}}$.After the first impact,the velocity becomes $v_1 = e \sqrt{2gh}$. The time taken

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(C) The time taken for the first drop is $t_0 = \sqrt{\frac{2h}{g}}$.After the first impact,the velocity becomes $v_1 = e \sqrt{2gh}$. The time taken for the first rebound is $t_1 = \frac{2v_1}{g} = 2e \sqrt{\frac{2h}{g}}$.Similarly,the time for the $n$-th rebound is $t_n = 2e^n \sqrt{\frac{2h}{g}}$.The total time $T$ is given by $T = t_0 + t_1 + t_2 + \dots = \sqrt{\frac{2h}{g}} [1 + 2(e + e^2 + e^3 + \dots)]$.Using the sum of an infinite geometric series $S = \frac{a}{1-r}$,where $a = e$ and $r = e$:$T = \sqrt{\frac{2h}{g}} [1 + 2(\frac{e}{1-e})]$.Given $h = 5 \ m$,$g = 10 \ m/s^2$,and $e = 1/2$:$T = \sqrt{\frac{2 	imes 5}{10}} [1 + 2(\frac{1/2}{1 - 1/2})] = 1 	imes [1 + 2(1)] = 3 \ s$.

$A$ ball is dropped from a height of $5 \ m$. The time after which the ball stops rebounding,if the coefficient of restitution between the ball and the ground is $e = 1/2$,is ................. $s$.

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