A small elastic ball falls from a height $h = 2.5 ext{ m}$ with zero initial velocity onto a smoot

Question

A small elastic ball falls from a height $h = 2.5 	ext{ m}$ with zero initial velocity onto a smooth inclined plane forming an angle $\alpha = 30^\circ$ with the horizontal. Having fallen the distance $h$, the ball rebounds elastically off the inclined plane. Find the distance (in meters) along the incline between the first and second impact points.

Professor PC Thomas Classes · Answer

Velocity before first impact $v_0 = \sqrt{2gh}$. Resolving axes along and perpendicular to incline: $u_x = v_0 \sin\alpha$, $u_y = v_0 \cos\alpha$ (elastic rebound). Accelerations: $a_x = g \sin\alpha$, $a_y = -g \cos\alpha$. Time of flight $T = \frac{2u_y}{|a_y|} = \frac{2v_0\cos\alpha}{g\cos\alpha} = \frac{2v_0}{g}$. Distance along incline $x = u_x T + \frac{1}{2}a_x T^2 = (v_0\sin\alpha)\left(\frac{2v_0}{g}ight) + \frac{1}{2}(g\sin\alpha)\left(\frac{4v_0^2}{g^2}ight) = \frac{4v_0^2}{g}\sin\alpha = 8h\sin\alpha$. Plugging $h=2.5, \alpha=30^\circ \Rightarrow x = 8(2.5)(0.5) = 10 	ext{ m}$.