A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is

If body is projected with velocity v$(v<v_e)$ then height up to which it will rise,$h=\frac{R}{\frac{v_e^2}{v^2}-1}$

$v=\frac{v_e}{2}$ $\therefore$(given) $h=\frac{R}{{\left(\frac{v_e}{v_e\text{}/\text{}2}\right)}^2-1}=\frac{R}{4-1}=\frac{R}{3}$

II method:

Total energy on surface=Total energy at a height

$$\begin{gathered} \frac{-GmM}{R}+\frac{1}{2}m{v}^2=\frac{-GmM}{R+h} \\ \Rightarrow \frac{1}{2}m{v}^2=\frac{-GmM}{R+h}+\frac{GmM}{R} \\ \Rightarrow v^2=2GM\left(\frac{1}{R}-\frac{1}{R+h}\right) \\ \Rightarrow v^2=2g{R}^2\left(\frac{R+h-R}{R(R+h)}\right)\left(\because g=\frac{GM}{R^2}\right) \\ \Rightarrow v^2=2gR\left(\frac{h}{(R+h)}\right)=\frac{2gh}{1+{\displaystyle \frac{h}{R}}} \end{gathered}$$

$$\begin{gathered} Now , {\left(\frac{v_e}{n}\right)}^2=\frac{2gh}{1+{\displaystyle \frac{h}{R}}} \\ \Rightarrow {\left(\frac{\sqrt{2gR}}{n}\right)}^2=\frac{2gh}{1+{\displaystyle \frac{h}{R}}} \\ \Rightarrow \frac{2gR}{n^2}=\frac{2gh}{1+{\displaystyle \frac{h}{R}}} \\ \Rightarrow \frac{R}{n^2}=\frac{h}{1+{\displaystyle \frac{h}{R}}} \\ \Rightarrow R\left(1+\frac{h}{R}\right)=n^{2}h \\ \Rightarrow \left(R+\frac{Rh}{R}\right)=n^{2}h \\ \Rightarrow \left(R+h\right)=n^{2}h \\ \Rightarrow h(n^2-1)=R \\ \Rightarrow h=\frac{\mathrm{R}}{({\mathrm{n}}^2-1)}=\frac{\mathrm{R}}{(2^2-1)}=\frac{R}{3} \end{gathered}$$