A body of mass m is projected horizontally just above the earth surface with velocity  $v_0=\sqrt{ngR}$, where R is the radius of earth. If maximum height attained by the body from the surface of earth is 2R, find the value of n.

From conservation of energy
 $$\begin{gathered} \frac{1}{2}m{v}_0^2-\frac{GMm}{R}=\frac{1}{2}m{v}^2-\frac{GMm}{R+h} \\ (V_0^2-V^2)=2\frac{GMh}{R(R+h)} \end{gathered}$$
 
 $V_1^2-V^2=\frac{4GM}{3R}$ ……. (1)
From conservation of angular momentum about centre of earth.
 $m{v}_{0}R=mv(R+h)$
$v=\frac{v_{0}R}{R+h}=\frac{v_0}{3}$  ……. (2)
 $$\begin{gathered} \therefore v_0^2-\frac{v_0^2}{9}=\frac{4}{3}\frac{GM}{R} \\ \frac{8}{9}n.g.R=\frac{4}{3}.gr \\ n=\frac{3}{2} \end{gathered}$$
 n = 1.50