A particle is to be projected horizontally with velocity `v` from a point `P`, which is `60 m` above the foot of a plane inclined at angle `45^(@)` with horizontal as shown in figure. The particle hits the plane perpendicularly at `A`. After rebound from inlined plane it again hits at `B`. If coefficient of restitution between particles and plane is `(1)/(sqrt(2))` then,

Options

A. `v = 20 m//s`
B. `v = 10 m//s`
C. `AB = 80sqrt(2)`m
D. `AB = 80 m`

Answer

A, C

Explanation

`{:(u_(x)^(1)=v cos 45^(0),a_(x)^(1)=-g sin 45^(0)),(u_(y)^(1)=-v sin 45^(0),a_(y)^(1)=-g cos 45^(0)):}`
At point
`v_(x)^(1) = 0 rArr 0 = v cos 45^(0) - g sin 45^(0)t, t = v//g`
`Tan 45^(0) = (h-(1)/(2)g t^(2))/(vt)`, `h - (1)/(2)g t^(2) = v t`
`h = (1)/(2)g[(v^(2))/(g^(2))]+(v^(2))/(g)`
`h = (v^(2))/(g) + (v^(2))/(2g) = (3v^(2))/(2g)`, `v = sqrt((2gh)/(3)) = 20 m//s`
`v'=` velocity just before collision `= sqrt(2)`
`v'' =` velocity after collision `=v`
`t_(AB) = (2v)/(g cos 45^(0)) = 2sqrt(2)(v)/(g) = 4sqrt(2)sec`
`AB = (1)/(2) xx 10 xx (1)/(sqrt(2)) xx 16 xx 2 = 80sqrt(2)_(m)`