The escape speed of a projectile on the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Escape velocity of a projectile from the Earth, $v_{\mathrm{esc}}=11.2 \mathrm{~km} / \mathrm{s}$
Projection velocity of the projectile, $v_{p}=3 v_{e s c}$
Mass of the projectile $=m$
Velocity of the projectile far away from the Earth $=v_{f}$
Total energy of the projectile on the Earth $=\frac{1}{2} m v_{p}^{2}-\frac{1}{2} m v_{\mathrm{cc}}^{2}$
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth $=\frac{1}{2} m v_{f}^{2}$
From the law of conservation of energy, we have
$\frac{1}{2} m v_{\mathrm{p}}^{2}-\frac{1}{2} m v_{\mathrm{cc}}^{2}=\frac{1}{2} m v_{\mathrm{f}}^{2}$
$v_{\mathrm{f}}=\sqrt{v_{\mathrm{p}}^{2}-v_{\mathrm{esc}}^{2}}$
$=\sqrt{\left(3 v_{\mathrm{esc}}\right)^{2}-\left(v_{\mathrm{ec}}\right)^{2}}$
$=\sqrt{8} v_{\mathrm{esc}}$
$=\sqrt{8} \times 11.2=31.68 \mathrm{~km} / \mathrm{s}$