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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 `:` A. `vsqrt((2h)/(g))`B. `2vsqrt((2h)/(g))`C. `3v sqrt((2h)/(g))`D. `4v sqrt((2h)/(g))` |
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Answer» Correct Answer - C 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 `)`. |
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