A ball falls freely from a height of 180 ,m o | vedclass

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(C) When a ball is dropped from height $h$, the time taken to reach the ground is $t_0 = \sqrt{\frac{2h}{g}}$ and the speed just before impact is $v_0

Vedclass · Accepted Answer

$\frac{50}{3} \,ms^{-1}, 10 \,ms^{-1}$

Vedclass · Answer

(C) When a ball is dropped from height $h$, the time taken to reach the ground is $t_0 = \sqrt{\frac{2h}{g}}$ and the speed just before impact is $v_0 = \sqrt{2gh}$.After the first collision, its speed becomes $v_1 = ev_0 = e\sqrt{2gh}$, where $e$ is the coefficient of restitution.The ball rises and stops at time $t_1 = \frac{v_1}{g}$. It then falls back to the ground, taking the same time $t_1$. Thus, the time between the first and second collision is $2t_1 = \frac{2v_1}{g}$.The total time $T$ before it ceases to rebound is:$T = t_0 + 2t_1 + 2t_2 + \dots = t_0 + \frac{2v_1}{g} + \frac{2v_2}{g} + \dots = \sqrt{\frac{2h}{g}} + \frac{2e\sqrt{2gh}}{g} + \frac{2e^2\sqrt{2gh}}{g} + \dots$$T = \sqrt{\frac{2h}{g}} [1 + 2e(1 + e + e^2 + \dots)] = \sqrt{\frac{2h}{g}} \left( \frac{1+e}{1-e} ight)$.Given $h = 180 \,m$, $g = 10 \,ms^{-2}$, and $e = 0.5$:$T = \sqrt{\frac{2 	imes 180}{10}} \left( \frac{1+0.5}{1-0.5} ight) = 6 	imes 3 = 18 \,s$.The total distance $H$ covered is:$H = h + 2h_1 + 2h_2 + \dots = h + 2(e^2h) + 2(e^4h) + \dots = h \left( \frac{1+e^2}{1-e^2} ight)$.$H = 180 \left( \frac{1 + 0.25}{1 - 0.25} ight) = 180 	imes \frac{1.25}{0.75} = 180 	imes \frac{5}{3} = 300 \,m$.Average speed $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{300}{18} = \frac{50}{3} \,ms^{-1}$.Average velocity $= \frac{	ext{Total displacement}}{	ext{Total time}} = \frac{180}{18} = 10 \,ms^{-1}$.

$A$ ball falls freely from a height of $180 \,m$ onto a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is $0.5$, the average speed and average velocity of the ball before it ceases to rebound are respectively (acceleration due to gravity $= 10 \,ms^{-2}$)

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