A body is projected up with a velocity $2$ times the escape velocity $v_{e}$ from the surface of the earth. The velocity of the body at a point far away the earth's attraction is

Let

$v$ be the speed of the body when it is at infinite distance from the center of the earth and $u$ be speed of projection of the body from the earth's surface.

According to law of conservation of mechanical energy,

$\begin{array}{l} \frac{1}{2} m u^{2}-\frac{G M m}{R}=\frac{1}{2} m v^{2}-0 \\ \\ v^{2}=u^{2}-\frac{2 G M}{R} \end{array}$

$\mathrm{v}^{2}=\mathrm{u}^{2}-\mathrm{v}_{\mathrm{e}}{ }^{2}\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{ \frac{2 G M}{R}}\right)$

Given that

$\mathrm{u}=2 \mathrm{~V}_{\mathrm{e}}$

So, $v=\sqrt{\left(2 V_{e}\right)^{2}-V_{e}^{2}}=\sqrt{3} V_{e}$