Q 11-03-172NEETNEET 2022Top questionEasy
A ball is projected with a velocity, $10\ \text{m s}^{-1}$, at an angle of $60°$ with the vertical direction. Its speed at the highest point of its trajectory will be
Answer: (C) $5\sqrt{3}\ \text{m s}^{-1}$
At the highest point only the horizontal component is left. The angle is measured from the vertical, so the horizontal component is $u\sin 60°$:
$$v = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics