A ball is thrown such that its horizontal range is exactly 4 3...

The horizontal range $R$ and maximum height $H$ of a projectile are given by:

$R = \dfrac{u^2 \sin(2\theta)}{g} = \dfrac{u^2 (2 \sin\theta \cos\theta)}{g}$

$H = \dfrac{u^2 \sin^2\theta}{2g}$

According to the given condition, $R = 4\sqrt{3} H$. Substituting the formulas:

$\dfrac{u^2 (2 \sin\theta \cos\theta)}{g} = 4\sqrt{3} \left( \dfrac{u^2 \sin^2\theta}{2g} \right)$

Cancelling $\dfrac{u^2}{g}$ from both sides:

$2 \sin\theta \cos\theta = 2\sqrt{3} \sin^2\theta$

Assuming $\theta \neq 0$, we can divide both sides by $2 \sin\theta \cos\theta$:

$1 = \sqrt{3} \tan\theta$

$\tan\theta = \dfrac{1}{\sqrt{3}}$

$\theta = 30^{\circ}$

Answer: $30^{\circ}$