Projection velocity of the projectile, \(v_{ p }=3 vesc\)

Mass of the projectile \(=m\)

Velocity of the projectile far away from the Earth \(=v_{ f }\)

Total energy of the projectile on the Earth \(=\frac{1}{2} m v_{ p }^{2}-\frac{1}{2} m v_{ ec }^{2}\)

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 \(\frac{1}{2} m v_{ p }^{2}-\frac{1}{2} m v_{ ec }^{2}=\frac{1}{2} m v_{ f }^{2}\)

\(v_{ f }=\sqrt{v_{ p }^{2}-v_{ cec }^{2}}\)

\(=\sqrt{\left(3 v_{ cec }\right)^{2}-\left(v_{ cec }\right)^{2}}\)

\(=\sqrt{8} v_{esc}\)

\(=\sqrt{8} \times 11.2=31.68 \;km / s\)