Q 11-03-034NEETJEE MainMedium
A ball is projected at $20$ m/s at $60°$ above the horizontal. After what time will its velocity be perpendicular to its initial velocity? ($g = 10\ \text{m/s}^2$)
Answer: (C) $\dfrac{4}{\sqrt{3}}$ s
Concept. $\vec v = \vec u + \vec gt$; perpendicular to $\vec u$ means $\vec u \cdot \vec v = 0$.
Step 1. Condition. $u^2 - ugt\sin\theta = 0 \Rightarrow t = \dfrac{u}{g\sin\theta}$.
Step 2. Substitute. $t = \dfrac{20}{10 \times \sqrt3/2} = \dfrac{4}{\sqrt3}$ s (after the top at $\sqrt3$ s, as it must be).
Answer. $\dfrac{4}{\sqrt3}$ s (Option C).
Solution by Sreeraj P, M.Sc Physics