A projectile of mass $\mathrm{m}$is fired from the surface of the earth at an angle $\alpha = {60}^{\circ }$from the vertical . The initial speed ${\nu }_0$ is equal to $\sqrt{\frac{G{M}_e}{R_e}}$. How does the projectile rise? Neglect air resistance and the earth's rotation.

 

Let $V$ be the speed of the projectile at highest point and $r_{max}$ its distance from the centre of the earth. Applying conservation of angular momentum and mechanical energy, 

$$\begin{gathered} m{v}_0\sin \alpha = mv{r}_{max} ......\left(i\right) \\ \frac{1}{2}m{v}_0^2 -\frac{G{M}_{e}m}{R_e} = \frac{1}{2} m{v}^2 - \frac{G{M}_{e}m}{r_{max}} .......\left(ii\right) \end{gathered}$$

Solving these two equations with given data we get, $r_{max} =\frac{3R_e}{2}$

or the maximum height $h_{max} = r_{max} - R_e = \frac{R_e}{2}$