A very naive Q

here v indicates the final velocity of the ball

so we have

iniitial velocity along the incline - vcos45=v/√2

intitial velocity erpendicular to it - vsin45=v/√2

further acceleration in direction of intitial velocity along incline= g/√2
and perpendicular to it= -g/√2

so let the ball return to the incline after time "t"
we have

v/√2 - g/√2t=-v/√2
√2v=gt/√2

t=2v/g

so let range =s

s=v/√2t+1/2(g/√2)t²

we already know t

damn, damn , damn I had calculated time time as v/g. Thanks rohan