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a body is thrown vetically up from the ground passes the height 10.2 m twice at an interval of 10s .what was initial velocity?

  • physics -

    The body spent 5.1 s going up above 10.2 m, and another 5.1 s going down. Its speed at that point was therefore

    v = 9.8*5.1 = 49.98 m/s @ y = 10.2 m_

    For the initial velocity Vo, use conservation of energy:
    Vo^2/2 - g*10.2 m = v^2/2
    Vo^2/2 = 1249 + 100
    Vo = 51.9 m/s

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