see Fig below

 

 

since collision is elastic e = 1 so velocity before and after collision along normal of the incline is same

let the velocity after collision be V at an angle # with the inclined plane

 

so vcos@ = Vsin# --------------------------1

also momentum along inclined plane is conserved so

vsin@ = Vcos# -------------------------------2

 

now after collinding with the plane the motion will be projectile so time taken for it to go up & come down is 2Vsin#/(gcos@)

 

so range is distance travelled by horizontal compenent in time 2Vsin#/(gcos@)

 

range(R) =  Vcos#(2Vsin#/(gcos@)) + 1/2*gsin@*[2Vsin#/(gcos@)]^2

             =  2Vsin#/(gcos@)*[Vcos# + sin@Vsin#/cos@] -----------3

from 1 & 2

Vsin# = vcos@

Vcos# = vsin@

putting these values in 3 we get

 

2vcos@/gcos@*[vsin@ + sin@vcos@/cos@]

= 2v/g[2vsin@]

= 4v^2sin@/g

 

hence the ball will hit the plane at a distance of 4v^2sin@/g down the incline