A particle is projected up a fixed inclined plane of inclination $30^\circ$, from the foot of the plane, with a speed of $20$ m/s. It strikes the inclined plane at right angles. The time of flight of the particle is ($g = 10\ \text{m/s}^2$)
Answer: (A) $\dfrac{8}{\sqrt7}$ s
Concept. Take axes along and perpendicular to the incline. Gravity has components $g\sin\theta$ down the incline and $g\cos\theta$ into it. Let the velocity make angle $\alpha$ with the incline. Striking at right angles means the velocity component along the incline is zero at landing.
Step 1. Time of flight. Perpendicular motion returns to the incline after
$$t = \frac{2u\sin\alpha}{g\cos\theta}$$
Step 2. Perpendicular-strike condition. Along the incline:
$$u\cos\alpha = (g\sin\theta)\,t = 2u\sin\alpha\tan\theta \;\Rightarrow\; \cot\alpha = 2\tan\theta = \frac{2}{\sqrt3}$$
So $\tan\alpha = \dfrac{\sqrt3}{2}$ and $\sin\alpha = \dfrac{\sqrt3}{\sqrt7}$.
Step 3. Substitute.
$$t = \frac{2(20)\frac{\sqrt3}{\sqrt7}}{10 \cdot \frac{\sqrt3}{2}} = \frac{8}{\sqrt7}\ \text{s} \approx 3.0\ \text{s}$$
Answer. $\dfrac{8}{\sqrt7}$ s (Option A).
Solution by Sreeraj P, M.Sc Physics