The velocity with which a projectile must be fired so that it...

For a projectile to leave the gravitational field of the earth, its kinetic energy would be at least equal to its potential energy. That is $\frac{1}{2}$m v_e^2=G $\frac{M m}{R} \Rightarrow$v_e=$\sqrt{\frac{2 G M}{R}}$where $G$ is the universal gravitational constant, $M$ is the mass of the earth and $R$ is the radius of the earth. Hence the escape velocity of a projectile is independent of its own mass.