Q 11-03-036NEETJEE MainEasy
A ball is thrown horizontally at $20$ m/s from the top of a tower $45$ m high. The angle its velocity makes with the horizontal when it hits the ground is ($g = 10\ \text{m/s}^2$)
Answer: (A) $\tan^{-1}\left(\dfrac{3}{2}\right)$
Concept. Horizontal velocity stays constant; vertical velocity grows as $gt$.
Step 1. Time of fall. $t = \sqrt{\dfrac{90}{10}} = 3$ s.
Step 2. Angle. $v_x = 20$, $v_y = 30$: $\tan\theta = \dfrac32$.
Answer. $\tan^{-1}\left(\dfrac32\right)$ (Option A).
Solution by Sreeraj P, M.Sc Physics