When a body is projected vertically upward with velocity v (not necessarily small compared to $v_e$), the maximum height is NOT simply v^$\frac{2}{2g}$.

Correct formula (using energy conservation): \frac{1}{2}$$mv^2 - GMm/R = -$\frac{GMm}{R+h}$ Solving: h = $v^2$*R / (2gR - $v^2$)

For v << $v_e$: h ≈ v^$\frac{2}{2g}$ (standard formula) For v = $v_e$: h -> infinity (escapes) For v = $v_e$/2: h = R/3

This is a common JEE numerical question.