Q. The equation of motion of a projectile are given by $x = 36\ t\ metre$ and $2y = 96\ t - 9.8\ t^{2}\ metre$. The angle of projection is
A) $\sin^{- 1}\left( \frac{4}{5} \right)$
B) $\sin^{- 1}\left( \frac{3}{5} \right)$
C) $\sin^{- 1}\left( \frac{4}{3} \right)$
D) $\sin^{- 1}\left( \frac{3}{4} \right)$
Explanation
(a) $x = 36t\ \therefore v_{x} = \frac{dx}{dt} = 36\ m/s$
$y = 48t - 4.9t^{2}\therefore v_{y} = 48 - 9.8t$
at $t = 0\ v_{x} = 36$ and $v_{y} = 48\ m/s$
So, angle of projection $\theta = \tan^{- 1}\left( \frac{v_{y}}{v_{x}} \right) = \tan^{- 1}\left( \frac{4}{3} \right)$
Or $\theta = \sin^{- 1}{(4/5)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.