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)

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