The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. Its angle of projection will be:

Question

Accepted Answer

1.  **Formulas:** 
    *   Maximum Height ($H$) is given by $H = \frac{v_0^2 \sin^2 	heta}{2g}$ .
    *   Horizontal Range ($R$) is given by $R = \frac{v_0^2 \sin 2	heta}{g} = \frac{2v_0^2 \sin 	heta \cos 	heta}{g}$ .
2.  **Given Condition:** $R = 4\sqrt{3} H$.
3.  **Substitution:**
    $\frac{2v_0^2 \sin 	heta \cos 	heta}{g} = 4\sqrt{3} \left( \frac{v_0^2 \sin^2 	heta}{2g} ight)$
4.  **Simplification:**
    Cancel common terms ($v_0^2$, $g$, and one $\sin 	heta$):
    $2 \cos 	heta = 4\sqrt{3} \left( \frac{\sin 	heta}{2} ight)$
    $2 \cos 	heta = 2\sqrt{3} \sin 	heta$
    $\frac{\sin 	heta}{\cos 	heta} = \frac{1}{\sqrt{3}}$
    $	an 	heta = \frac{1}{\sqrt{3}}$
5.  **Calculate Angle:**
    $	heta = 30^{\circ}$.

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