A body is projected vertically upwards from earth's surface with velocity \(2 \mathrm{v}_{\mathrm{e}}\), where \(\mathrm{v}_{\mathrm{e}}\) is escape velocity from earth's surface. The velocity when body escapes the gravitational pull is

The kinetic energy given to the body \(=\mathrm{k} =\frac{1}{2} \mathrm{~m}(2 \mathrm{Ve})^{2} \)
\( =\frac{1}{2}\left(4 \mathrm{mV}_{\mathrm{o}}^{2}\right)\)
The kinetic energy required to escape from the earth's gravitational field \(=\mathbf{K}^{\prime}=\frac{1}{2} \mathbf{m V}_{0}^{2}\)
If its velocity is \(\mathrm{V}\) after escaping from the earth's gravitational field then its kinetic energy will be
\(
\begin{aligned}
\frac{1}{2} \mathrm{mV}^{2} &=\frac{1}{2}\left(4 \mathrm{mV}_{\mathrm{e}}^{2}\right)-\frac{1}{2} \mathrm{mV}_{\mathrm{e}}^{2} \\
&=\frac{1}{2}\left(3 \mathrm{mV}_{\mathrm{e}}^{2}\right) \\
\therefore \mathrm{V}^{2} &=3 \mathrm{~V}_{\mathrm{e}}^{2} \\
\therefore \mathrm{V} &=\sqrt{3} \mathrm{~V}_{\mathrm{e}}
\end{aligned}
\)