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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    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) \[v\sqrt{\frac{2h}{g}}\]  

    B)        \[2v\sqrt{\frac{2h}{g}}\]

    C) \[4v\sqrt{\frac{2h}{g}}\]

    D)        \[3v\sqrt{\frac{2h}{g}}\]

    Correct Answer: D

    Solution :

    [d] Solve in the reference frame fixed to the wall. Before collision, velocity of ball = 3v towards it. \[\therefore \] After elastic collision of ball = 3v away from it Time of flight \[=\sqrt{\frac{2h}{g}}\] \[\therefore \] Distance between wall and ball \[=3v\,\sqrt{\frac{2h}{g}}.\]


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