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(A) Let the ball be projected vertically downward with velocity $v$ from height $h = 20 \ m$.Total mechanical energy at the point of projection $A$ is

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$20$

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(A) Let the ball be projected vertically downward with velocity $v$ from height $h = 20 \ m$.Total mechanical energy at the point of projection $A$ is $E_i = \frac{1}{2}mv^2 + mgh$.During the collision with the floor,the ball loses $50\%$ of its energy. The remaining energy is $E_f = 0.5 	imes E_i = 0.5 \left( \frac{1}{2}mv^2 + mgh ight)$.After the collision,the ball rebounds to the same height $h$. At this maximum height,its velocity is zero,so its total energy is purely potential energy: $E_{final} = mgh$.Equating the energy after collision to the energy required to reach height $h$: $0.5 \left( \frac{1}{2}mv^2 + mgh ight) = mgh$.Dividing both sides by $m$ and simplifying: $\frac{1}{4}v^2 + \frac{1}{2}gh = gh$.$\frac{1}{4}v^2 = \frac{1}{2}gh$.$v^2 = 2gh$.$v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 20} = \sqrt{400} = 20 \ m/s$.

$A$ ball is projected vertically downward with an initial velocity from a height of $20 \ m$ onto a horizontal floor. During the impact,it loses $50\%$ of its energy and rebounds to the same height. The initial velocity of its projection is ............ $m/s$.

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