Escape velocity of a projectile from the Earth, $v_{ esc }=11.2 km / s$

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

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

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

$=\sqrt{8} v_{esc}$

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