A ball collides elastically with a massive wall moving towards it with a velocity of `v` as shown. The collision occurs at a height of `h` above ground level and the velocity of the ball just before collision is `2v` in horizontal direction. The distance between the foot of the wall and the point on the ground where the ball lands, at the instant the ball lands, will be `:`

Options

A. `vsqrt((2h)/(g))`
B. `2vsqrt((2h)/(g))`
C. `3v sqrt((2h)/(g))`
D. `4v sqrt((2h)/(g))`

Answer

C

Explanation

Solve in the reference frame fixed to the wall.
Before collision, velocity of ball `=3v` towards it .
`:.` After elastic collision of ball`=3v` away from it
Time of flight `=sqrt((2h)/(g))`
`:.` distance between wall and ball
`=3v. sqrt((2h)/(g)).`
`(` Here no pseudo force is applied since the wall keeps on moving with constant velocity `w.r.t.` ground, it being very heavy `)`.