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A ball is thrown vertically down from a height of $40 \mathrm{~m}$ from the ground with an initial velocity $v$. The ball hits the ground, loses $\frac{1}{3}$ rd of its total mechanical energy and rebounds back to the same height. If the acceleration due to gravity is $10 \mathrm{~ms}^{-2}$, then the value of $v$ is
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Verified Answer
The correct answer is:
$20 \mathrm{~ms}^{-1}$
As per question, the ball hits the ground loses $\frac{1}{3} \mathrm{rd}$ of its total mechanical energy and rebounds back to the same height.
$\begin{aligned}
& \text { i.e., } \frac{2}{3}\left(\frac{1}{2} m v^2+m g h\right)=m g h \\
& \Rightarrow \quad \frac{2}{3}\left(\frac{1}{2} v^2+g h\right)=g h \\
& \Rightarrow \quad \frac{v^2}{3}=g h-\frac{2 g h}{3} \\
& \Rightarrow \quad \frac{v^2}{3}=\frac{g h}{3} \Rightarrow v=\sqrt{g h} \\
& \therefore \quad v=\sqrt{400}=20 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\begin{aligned}
& \text { i.e., } \frac{2}{3}\left(\frac{1}{2} m v^2+m g h\right)=m g h \\
& \Rightarrow \quad \frac{2}{3}\left(\frac{1}{2} v^2+g h\right)=g h \\
& \Rightarrow \quad \frac{v^2}{3}=g h-\frac{2 g h}{3} \\
& \Rightarrow \quad \frac{v^2}{3}=\frac{g h}{3} \Rightarrow v=\sqrt{g h} \\
& \therefore \quad v=\sqrt{400}=20 \mathrm{~m} / \mathrm{s}
\end{aligned}$
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