Physics - W.E.P. Question with Solution | TestHub

Let a particle of mass $m$ is thrown vertically upward with velocity $v$. Let its velocity be $v^{\prime }$ at a height $3R$ from the surface of the earth. According to the conservation of energy, $\frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^{\prime 2}-\frac{GMm}{(R+3R)}$ $v^2-\frac{2GM}{R}=v^{\prime 2}-\frac{2GM}{4R}$ $v^2-v^{\prime 2}=\frac{-2GM}{4R}+\frac{2GM}{R}=\frac{6}{4}\frac{GM}{R}$ $v^{\prime 2}=v^2-\frac{6}{4}\frac{GM}{R}$ $v^{\prime 2}=\sqrt{(11.2{)}^2-\frac{3}{2}gR}\left(\because g=\frac{GM}{R^2}\right)$ $=\sqrt{(11.2{)}^2-\frac{3}{2}\times \frac{9.8\times 6400}{1000}}$ $=\sqrt{(11.2{)}^2-(94.08)}$ $=5.6\mathrm{km}{\mathrm{s}}^{-1}$