A projectile is projected with velocity k.v_e in vertically upward direction from the ground into the space. (ve is escape velocity and k < 1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be (R = radius of earth)

(c) R/(1 -  k²⁾

From the law of conservation of energy Difference in potential energy between ground and maximum height = Kinetic energy at the point of projection :

\(\frac{mgh}{1+\frac{h}{R}} = \frac{1}{2} m (k.v_e)^2\)

   = \(\frac{1}{2} m k^2 (\sqrt{2 g R})^2\) ... [As \(v_e = \sqrt{2gR}\)

By solving height from the surface of earth : 

\(h = \frac{Rk^2}{1-k^2}\)

So height from the centre of earth

\(r = R + h = R + \frac{R}{1 - k^2}\)