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A ball of mass $0.2 \mathrm{~kg}$ is thrown vertically down from a height of $10 \mathrm{~m}$. It collides with the floor and loses $50 \%$ of its energy and then rises back to the same height. The value of its initial velocity is
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$14 \mathrm{~ms}^{-1}$
Given, mass of the ball, $m=0.2 \mathrm{~kg}$
Height from the surface of floor, $h=10 \mathrm{~m}$
Let initial velocity be $u$.
Total energy as initial point $\mathrm{TE}=\frac{1}{2} m v^2+m g h$
After collision, remaining energy
$=\frac{\frac{1}{2} m u^2+m g h}{2}$
With this remaining energy the ball bounces upto height $h$
Therefore, $\frac{\frac{1}{2} m u^2+m g h}{2}=m g h$
$\begin{aligned} & \Rightarrow \frac{1}{2} m u^2+m g h=2 m g h \\ & \Rightarrow \begin{aligned} u^2 & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 10}=14 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$
Height from the surface of floor, $h=10 \mathrm{~m}$
Let initial velocity be $u$.
Total energy as initial point $\mathrm{TE}=\frac{1}{2} m v^2+m g h$
After collision, remaining energy
$=\frac{\frac{1}{2} m u^2+m g h}{2}$
With this remaining energy the ball bounces upto height $h$
Therefore, $\frac{\frac{1}{2} m u^2+m g h}{2}=m g h$
$\begin{aligned} & \Rightarrow \frac{1}{2} m u^2+m g h=2 m g h \\ & \Rightarrow \begin{aligned} u^2 & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 10}=14 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$
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