If we rotate the coordinate system so that the ramp is horizontal, then the free fall acceleration will have two components one downward ${\mathrm{a}}_{\mathrm{y}}=-\mathrm{ }\mathrm{g}\mathrm{ }\cos \mathrm{\theta }$ and one horizontal ${\mathrm{a}}_{\mathrm{x}}=\mathrm{g}\mathrm{ }\sin \mathrm{\theta . }$In this frame, the initial velocity will have components given by ${\mathrm{V}}_{\mathrm{y}}={\mathrm{V}}_0\cos \mathrm{\theta }$ & ${\mathrm{V}}_{\mathrm{x}}={\mathrm{V}}_0\mathrm{ }\sin \mathrm{\theta }$ time for each bounce is given by ${\mathrm{t}}_{\mathrm{b}}=\frac{2{\mathrm{V}}_y}{-ay}=\frac{2{\mathrm{V}}_0}{g}$

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

given these equation

${\mathrm{d}}_{12}={\mathrm{V}}_0\mathrm{ }\sin \mathrm{\theta }\left(\frac{2V_0}{g}\right)\mathrm{ }+\mathrm{ }0.5\mathrm{g}\mathrm{ }\sin \mathrm{\theta }\mathrm{ }{\left(\frac{2V_0}{g}\right)}^2$

$=\mathrm{ }\frac{4V_0^2\sin \mathrm{\theta }}{g}$

${\mathrm{d}}_{13}={\mathrm{V}}_0\sin \mathrm{\theta }\left(\frac{4V_0}{g}\right)\mathrm{ }+\mathrm{ }0.5\mathrm{ }\mathrm{g}\mathrm{ }\sin \mathrm{\theta }{\left(\frac{4V_0}{g}\right)}^2$

$=\mathrm{ }\frac{12{V_0}^2\sin \mathrm{\theta }}{g}$

Since ${\mathrm{d}}_{23}={\mathrm{d}}_{13}-\mathrm{ }{\mathrm{d}}_{12}=\mathrm{ }\frac{8{V_0}^2\sin \mathrm{\theta }}{g}$

$\frac{d_{12}}{d_{23}}=\frac{1}{2}$