Q. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
A) $\theta = \tan^{- 1}\left( \frac{1}{4} \right)$
B) $\theta = \tan^{- 1}(4)$
C) $\theta = \tan^{- 1}(2)$
D) $\theta = 45{^\circ}$
Explanation
(b) Horizontal range
$R = \frac{u^{2}\sin{2\theta}}{g}$ ...(i)
Maximum height
$H = \frac{u^{2}\sin^{2}\theta}{2g}$ ...(ii)
Here (i)=(ii)
$\frac{u^{2}\sin{2\theta}}{g} = \frac{u^{2}\sin^{2}\theta}{2g}$
$2\cos\theta = \frac{\sin\theta}{2}$
$\theta = \tan^{- 1}{(4)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.