A plane is inclined at an angle $\alpha = 30^\circ$ with respect to the horizontal. A particle is projected up the plane with a speed $u = 2\ \text{m s}^{-1}$, from the base of the plane, making an angle $\theta = 15^\circ$ with respect to the plane. The distance from the base, at which the particle hits the plane is close to: (Take $g = 10\ \text{m s}^{-2}$)
Answer: (A) $20\ \text{cm}$
Take axes along and perpendicular to the incline. Perpendicular to the plane: initial velocity $u\sin\theta$, acceleration $g\cos\alpha$, so the time of flight is
$$T = \frac{2u\sin\theta}{g\cos\alpha}$$
Along the plane: $R = u\cos\theta\,T - \tfrac12g\sin\alpha\,T^2$, which simplifies to
$$R = \frac{2u^2\sin\theta\cos(\theta+\alpha)}{g\cos^2\alpha} = \frac{2\times4\times\sin15^\circ\cos45^\circ}{10\times0.75} = \frac{8\times0.259\times0.707}{7.5} \approx 0.20\ \text{m}$$
So the particle lands about $20\ \text{cm}$ from the base.
Solution by Sreeraj P, M.Sc Physics