Explanation

(d) Let $u$ be initial velocity of projection at angle $\theta$ with the horizontal. Then, horizontal range,
$R = \frac{u^{2}\sin{2\theta}}{g}$
$and\ maximum\ height\ H = \frac{u^{2}\sin{2\theta}}{2g}$
$Given,\ \ \ R = 4\sqrt{3}H$
$\therefore\ \ \frac{u^{2}\sin{2\theta}}{g} = 4\sqrt{3} \bullet \frac{u^{2}\sin^{2}\theta}{2g}$
$\therefore\ 2\sin{\theta\cos{\theta = 2\sqrt{3}\ \sin^{2}\theta}}$
$or\ \ \frac{\cos\theta}{\sin\theta} = \sqrt{3}$
$or\cot{\theta = \sqrt{3} = \cot{30{^\circ}}}$