Explanation

(b) Let the angle of projection be $\alpha$.
$\therefore Range,\ R = \frac{u^{2}\sin{2\alpha}}{g}$
$and\ maximum\ height\ H = \frac{u^{2}\sin^{2}\alpha}{2g}$
Now, it is given that,
$(Range)^{2} = 48(maximum\ height)^{2}$
$\therefore\ \left( \frac{u^{2}\sin{2\alpha}}{g} \right)^{2} = 48\left( \frac{u^{2}\sin^{2}\alpha}{2g} \right)^{2}$
$or\ \frac{u^{2}\sin{2\alpha}}{g} = 4\sqrt{3}\left( \frac{u^{2}\sin^{2}\alpha}{2g} \right)$
$or\ \frac{2\sin{\alpha\cos\alpha}}{4\sqrt{3}} = \frac{\sin^{2}\alpha}{2}$
$or\ \tan{\alpha = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}}$
$\therefore\ \alpha = 30{^\circ}$