Question

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

• A. $45^{\circ}$
• B. $60^{\circ}$
• C. $90^{\circ}$
• D. $30^{\circ}$

Answer 1

Answer: (D)

$30^{\circ}$

Answer 2

Let u be initial velocity of projection at angle $\theta$ with the horizontal. Then, horizontal range, $R=4 \sqrt{3}$ $\therefore \frac{u^{2} \sin 2 \theta}{g}=4 \sqrt{3} \cdot \frac{u^{2} \sin ^{2} \theta}{2 g}$ or $2 \sin \theta \cos \theta=2 \sqrt{3} \sin ^{2} \theta$ or $\frac{\cos \theta}{\sin \theta}=\sqrt{3}$ $\cot \theta=\sqrt{3}$ $\theta =30^{\circ}$.