Q. The horizontal range of a projectile $4\sqrt{3}$ times the maximum height achieved by it, then the angle of projection is
A) $30{^\circ}$
B) $45{^\circ}$
C) $60{^\circ}$
D) $90{^\circ}$
Explanation
(a) $\frac{u^{2}\sin{2\theta}}{g} = 4\sqrt{3} \times \frac{u^{2}\sin\theta}{2g}$
or $\frac{u^{2}}{g}2\sin\theta\cos\theta = 2\sqrt{3}\frac{u^{2}}{g}\sin^{2}\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.