A particle is projected vertically upwards from the surface of the Earth with a velocity equal to half the escape velocity. The maximum height reached by the planet is:

Solution

1. Escape velocity $v_e=\sqrt{2GM/R}$.
2. Initial energy $$\begin{array}{rl}{E}_i& =1/2m(v_e/2{)}^2-GMm/R\\ & =1/2m(GM/2R)-GMm/R\\ & =-3GMm/4R\end{array}$$.
3. At max height $h$, $E_f=-GMm/(R+h)$.
4. Equating $E_i=E_f$, we get $3/4R = 1/(R+h)$.
5. Solving for $h$, $3R + 3h = 4R$, so $h=R/3$.
6. Hence the answer is (B).