If $h_1,h_2$ and $h_3$ are the displacements of object along the incline plane between $1st-2nd,2nd-3rd$ and $3rd-4th$ impact, then $h_1:h_2:h_3$.

R.E.F image

Suppose the object coincides at a velocity

V in the $1st$ impact.

Considering t challenges a projective motion the

velocity before the $2^{nd}$concision is V as were.

Suppose, the coefficient of restitution $=e$

Where, $e<1$ (for non - elastic concision)

The velocity after $2^{nd}$ concision is ev

and $3^{rd}$ collision is $e^{2}v$

Let the angle of inclined power be $\theta$

Vertical component of velocity after

$1st$ concision is $vcos\theta$

$2^{nd}$ concision is $evcos\theta$

$3^{rd}$ concision is $e^{2}vcos\theta$

$\frac{1}{2}m(vcos\theta {)}^2=mg{h}_1$

$h_1=\frac{(vcos\theta {)}^2}{2g}$

$h_2=\frac{(evcos\theta {)}^2}{2g}$

$h_3=\frac{(e^{2}vcos\theta {)}^2}{2g}$

$\therefore h_1:h_2:h_3=1:e^2:e^4$