A particle is thrown vertically upwards with velocity 11.2 ~km ~s^...

Let a particle of mass $m$ is thrown vertically upward with velocity $v$. Let its velocity be $v^{\prime}$ at a height $3 R$ from the surface of the earth. According to the conservation of energy, $\frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m v^{\prime 2}-\frac{G M m}{(R+3 R)}$ $v^2-\frac{2 G M}{R}=v^{\prime 2}-\frac{2 G M}{4 R}$ $v^2-v^{\prime 2}=\frac{-2 G M}{4 R}+\frac{2 G M}{R}=\frac{6}{4} \frac{G M}{R}$ $v^{\prime 2}=v^2-\frac{6}{4} \frac{G M}{R}$ $v^{\prime 2}=\sqrt{(11.2)^2-\frac{3}{2} g R} \quad\left(\because g=\frac{G M}{R^2}\right)$=\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}