Q. The equation of a projectile is $y = \sqrt{3}x - \frac{gx^{2}}{2}$. The angle of projection is given by
A) $\tan\theta = \frac{1}{\sqrt{3}}$
B) $\tan\theta = \sqrt{3}$
C) $\frac{\pi}{2}$
D) Zero
Explanation
(b) Computing the given equation with
$y = x\tan\theta - \frac{gx^{2}}{2v^{2}\cos^{2}\theta}$, we get
$\tan\theta = \sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.