A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10 sec. Its initial velocity was (g = 10 m/s2)

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  1. In this problem, it takes 5 seconds for the body to rise from 10.2 m elevation to the maximum height, and another 5 seconds to fall back to 10.2 m elevation.

    The speed at 10.2 m height is then

    V(10.2m) = g* 5 s = 50 m/s

    The initial velocity Vo is whatever allows the body to have a speed of 50 m/s when the elevation is 10.2 m

    Use conservation of energy.

    M g *10.2 m = (M/2)[Vo^2 - 50^2]

    M cancels out. Solve for Vo.

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  2. mein gawar hu mujhe nahi ata

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