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A body of mass $m$ is dropped from a height $\frac{R}{2}$, from the surface of earth where $R$ is radius of earth. Its speed when it will hit the earth's surface is $\left(v_e=\right.$ escape $v$ elocity from earth' surface)
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The correct answer is:
$\frac{V_e}{\sqrt{3}}$
Considering the total energy conservation:
$\begin{aligned}
& -\frac{G M m}{R+\frac{R}{2}}=-\frac{G M m}{R}+\frac{1}{2} m v^2 \\
& \Rightarrow v=\sqrt{\frac{2 G M}{3 R}}=\frac{v_e}{\sqrt{3}}
\end{aligned}$
$\begin{aligned}
& -\frac{G M m}{R+\frac{R}{2}}=-\frac{G M m}{R}+\frac{1}{2} m v^2 \\
& \Rightarrow v=\sqrt{\frac{2 G M}{3 R}}=\frac{v_e}{\sqrt{3}}
\end{aligned}$
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