Physics - Gravitation Question with Solution | TestHub

Here, maximum height attained by a projectile $$h=\frac{v^{2}R}{2gR-v^2}...(i)$$ Velocity of body $=$ half the escape velocity i.e., $v=\frac{v_e}{2}$ or $v=\frac{\sqrt{2gR}}{2}\Rightarrow v^2=\frac{2gR}{4}\Rightarrow v^2=\left(\frac{gR}{2}\right)$ Now, putting value of $v^2$ in Eq. (i), we get Height, $h=\frac{\frac{gR}{2}\cdot R}{2gR-\frac{gR}{2}}=\frac{\frac{g{R}^2}{2}}{\frac{3gR}{2}}=\frac{R}{3}$