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) $\tan^{-1}\left(\dfrac{3}{2}\right)$
Time of fall: $t = \sqrt{\dfrac{2h}{g}} = \sqrt{9} = 3$ s.
At impact $v_x = 20$ m/s and $v_y = gt = 30$ m/s:
$$\tan\theta = \frac{30}{20} = \frac{3}{2}$$
Solution by Sreeraj P, M.Sc Physics