An elastic ball is dropped on a long inclined plane. If bounces,...

If we rotate the coordinate system so that the ramp is horizontal, then the free fall acceleration will have two components one downward ${a}_{y} = - gcos \theta$ and one horizontal ${a}_{x} = gsin \theta .$ In this frame, the initial velocity will have components given by ${V}_{y} = {V}_{0}cos \theta$ & ${V}_{x} = {V}_{0}sin \theta$ time for each bounce is given by ${t}_{b} = \dfrac{2{V}_{y}}{- ay} = \dfrac{2{V}_{0}}{g}$

The horizontal displacement $\Deltax = {V}_{x}t + 0.5{a}_{x}t^{2}$

given these equation

${d}_{12} = {V}_{0}sin \theta\left(\dfrac{2{V}_{0}}{g}\right) + 0.5gsin \theta{\left(\dfrac{2{V}_{0}}{g}\right)}^{2}$

$= \dfrac{4{V}_{0}^{2}sin \theta}{g}$

${d}_{13} = {V}_{0}sin \theta\left(\dfrac{4{V}_{0}}{g}\right) + 0.5gsin \theta{\left(\dfrac{4{V}_{0}}{g}\right)}^{2}$

$= \dfrac{12{{V}_{0}}^{2}sin \theta}{g}$

Since ${d}_{23} = {d}_{13} - {d}_{12} = \dfrac{8{{V}_{0}}^{2}sin \theta}{g}$

$\dfrac{{d}_{12}}{{d}_{23}} = \dfrac{1}{2}$