`h=5m `
`theta=45^0`
e=(3/4)`
here thevelocity with which it wold strike
`=v=sqt(2gxx5)=10m/sec`
After colision let ilt makes an angle `beta` with horizontal. The horizontal components of velocity ` 10 cos45^0` will remain unchanged and velocity in the perpendicular direction to the plane after wilisine
`v_1=exx10sin45^0`
`=(3/4xx10xx1/sqrt(2)`
`=(3.75)sqrt2m/sec`
similarly `v_2=5sqrt2`m/sec
`u=sqt(v_2^2+v_1^2)`
`=sqrt(50+28.125)`
`=sqrt(78.125)`
`=8.83m/sec`
Angle of relfection from the wall
`beta=tan^-1((3.75sqrt2))/(5sqrt2)`
`=tan^-1(3/4)=37^0
`rarr Angle of projection `alpha=90-(theta+beta)`
`90-(45+37^0)=8^0`
Let the distance where it falls =L
`rarr x=Lcostheta`
Angle of projection `(alpha)=-8^0`
`y=xtaN/Alpha-(gx^2sec^2alpha)/(2u^2)`
`rarr -Lsintheta=Lcosthetaxxtan8^0
` rarr -g/2 (L^2cos^2thetasec^28^0)/((8.83)^2)
solving the above equation we get
`L=18.5m
