There is a crater of depth R/100 on the surface of the moon (radius R). A Projectile is fired vertically upward from the crater with velocity, which is equal to the escape velocity v from the surface of the moon. Find the maximum height attained by the projectile.

Speed of particle at A,
v_A = escape velocity on the surface of earth $=\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}$
At highest point B,  v_B = 0
Applying conservation of mechanical energy, decrease in kinetic energy = increase in gravitational potential energy
$=\frac{1}{2}{\mathrm{mv}}_{\mathrm{A}}^2={\mathrm{U}}_{\mathrm{B}}-{\mathrm{U}}_{\mathrm{A}}=\mathrm{m}\left({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}\right)$
$\begin{array}{l}\frac{{\mathrm{v}}_{\mathrm{A}}^2}{2}={\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}\\ \frac{\mathrm{GM}}{\mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}-\left[\frac{-\mathrm{GM}}{{\mathrm{R}}^3}\left[\left(1.5{\mathrm{R}}^2\right)-0.5{\left(\mathrm{R}-\frac{\mathrm{R}}{100}\right)}^2\right]\right]\\ \frac{1}{\mathrm{R}}=-\frac{1}{\mathrm{R}+\mathrm{h}}+\frac{3}{2\mathrm{R}}-\left(\frac{1}{2}\right){\left(\frac{99}{100}\right)}^2\cdot \frac{1}{\mathrm{R}}\end{array}$
Solving this equation, we get h = 99.5R