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  1. A ball is thrown vertically upward from the ground. It crosses a point at the height of 25 m twice at an interval of 4 seconds. The ball was thrown with the velocity of

    2. A.

     18 m/s
    B.

     25 m/s
    C.

     30 m/s
    D.

     36 m/s

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    Answer : Option C

    Explanation :

    The interval between object pass the same point is 4 sec.
    That means in 2 sec, object reaches the top and in next 2 sec, it again reaches the same point.
    By the info given, we can find the velocity of that point by using : v=u+(-gt)
    0= u - 10×2
    u = 20
    Then the velocity at 25m is 20m/sec.
    so the initial velocity is
    −v2 = u2 + (−2gh)
    400 = u2 − 2x 10 x 25
    u2 = 900
    => u = 30 m/s

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