A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle $\alpha ={37}^o$ with the horizontal. Having fallen the distance h = 1m, the ball rebounds elastically of the inclined plane. Find the distance (in m) moved by the ball along the plane between second and fourth impact point. The plane is enough long.

Since the collision is elastic, the velocity perpendicular to the incline is reversed just after the collision, thus the time of flight between any of the collisions remains same.
$T=\frac{2usin\theta }{gcos\alpha }=\frac{2usin{53}^0}{gcos{37}^0}=\frac{2\sqrt{2gh}}{g}{t}_2=\frac{2\sqrt{2gh}}{g}and{t}_4=\frac{6\sqrt{2gh}}{g}s=(u_x{t}_4+\frac{1}{2}{a}_x{t_4}^2)-(u_x{t}_2+\frac{1}{2}{a}_x{t_2}^2)\Rightarrow s=\frac{3\sqrt{2gh}}{5}\times \frac{6\sqrt{2gh}}{g}+\frac{gsin{37}^0}{2}\times {\left(\frac{6\sqrt{2gh}}{g}\right)}^2-(\frac{3\sqrt{2gh}}{5}\times \frac{2\sqrt{2gh}}{g}+\frac{gsin{37}^0}{2}\times {\left(\frac{2\sqrt{2gh}}{g}\right)}^2)\Rightarrow s=\frac{112h}{5}$