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
$\vec{v} = \vec{u} + \vec{g}t$. Perpendicular to $\vec{u}$ means $\vec{u}\cdot\vec{v} = 0$:
$$u^2 + (\vec{u}\cdot\vec{g})t = 0 \;\Rightarrow\; u^2 - ugt\sin\theta = 0 \;\Rightarrow\; t = \frac{u}{g\sin\theta}$$
$$t = \frac{20}{10 \times \sqrt{3}/2} = \frac{4}{\sqrt{3}}\ \text{s} \approx 2.3\ \text{s}$$
This is after the ball has passed its highest point ($\sqrt{3}$ s), as it must be.
Solution by Sreeraj P, M.Sc Physics