Explanation

(d) We know that if two stones have same horizontal range, then this implies that both are projected at $\theta$ and $90{^\circ} - \theta.$
$Given,\ \theta = \frac{\pi}{3} = 60{^\circ}$
$\therefore\ \ 90{^\circ} - \theta = 90{^\circ} - 60{^\circ} = 30{^\circ}$
For first stone,
$Maximum\ height = 102 = \frac{u^{2}\sin^{2}60{^\circ}}{2g}$
For second stone,
$Maximum\ height,\ h = \frac{u^{2}\sin^{2}30{^\circ}}{2g}$
$\therefore\ \frac{h}{102} = \frac{\sin^{2}30{^\circ}}{\sin^{2}60{^\circ}} = \frac{(1/2)^{2}}{\left( \sqrt{3}/2 \right)^{2}}$
$or\ h = 102 \times \frac{1/4}{3/4} = 34\ m$