A ball is falling freely from a height. When | vedclass

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(C) Let the mass of the ball be $m$. When the ball is at a height $h = 10 ~m$,its velocity is $v_0$. The kinetic energy at this point is $K_1 = \frac{

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$14$

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(C) Let the mass of the ball be $m$. When the ball is at a height $h = 10 ~m$,its velocity is $v_0$. The kinetic energy at this point is $K_1 = \frac{1}{2} m v_0^2$.Just before hitting the ground,the ball falls an additional $10 ~m$. By the work-energy theorem,the kinetic energy just before impact is $K_{impact} = K_1 + mgh = \frac{1}{2} m v_0^2 + mg(10)$.After the collision,the ball loses $50 \%$ of its energy,so the kinetic energy just after impact is $K_{after} = 0.5 	imes K_{impact} = 0.5 	imes (\frac{1}{2} m v_0^2 + 10mg)$.The ball rises back to a height of $10 ~m$,meaning its potential energy at the peak is $mgh = 10mg$. Since the kinetic energy is zero at the peak,by conservation of energy,$K_{after} = 10mg$.Equating the two: $0.5 	imes (\frac{1}{2} m v_0^2 + 10mg) = 10mg$.$\frac{1}{2} m v_0^2 + 10mg = 20mg$.$\frac{1}{2} m v_0^2 = 10mg$.$v_0^2 = 20g$.Using $g = 10 ~m/s^2$,$v_0^2 = 200$,so $v_0 = \sqrt{200} \approx 14.14 ~m/s$. Given the options,$14 ~m/s$ is the correct choice.

$A$ ball is falling freely from a height. When it reaches $10 ~m$ height from the ground,its velocity is $v_0$. It collides with the ground,loses $50 \%$ of its energy,and rises back to a height of $10 ~m$. Then the velocity $v_0$ is (in $~m/s$)

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