A body is projected from earth's with thrice the escape velocity from the surface of the earth. What will be its velocity when it will escape the gravitational pull?

Energy required to escape the earth's gravitational field is \(\frac{1}{2} \mathrm{mV}_e^2\)
Energy given to the body is \(=\frac{1}{2} \mathrm{~m}\left(3 \mathrm{~V}_{\mathrm{e}}\right)^2\)
\(=\frac{9}{2} \mathrm{mV}_{\mathrm{e}}^2\)
\(\therefore\) If \(\mathrm{V}\) is velocity of the body when it has escaped from earth's gravitational field then
\(
\begin{aligned}
& \frac{1}{2} \mathrm{mV}^2=\frac{9}{2} \mathrm{mV}_{\mathrm{e}}^2-\frac{1}{2} \mathrm{mV}_{\mathrm{e}}^2 \\
& \therefore \frac{1}{2} \mathrm{mV}^2=4 \mathrm{mV}_{\mathrm{e}}^2 \\
& \therefore \mathrm{V}^2=8 \mathrm{~V}_{\mathrm{e}}^2 \\
& \mathrm{~V}=2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}
\end{aligned}
\)