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5 June, 18:13

A stone is dropped from a 75-m - high building. When this stone has dropped 15 m, a second stone is thrown downward with an initial velocity such that the two stones hit the ground at the same time. What was the initial velocity of the second stone?

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  1. 5 June, 21:58
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    The initial velocity of the second stone is 34.3 m/s

    Explanation:

    The distance height covered by the second stone will be

    75m - 15m = 60 m

    Since the two stones hit the ground at the same time.

    Using third equation of motion, which state that;

    V^2 = U^2 - 2gH

    Where

    U = Initial velocity

    V = final velocity

    H = 60 m

    g = 9.81 m/s^2

    As the stone reaching the ground, that is, coming to rest, V = 0. Therefore,

    Substitute all the parameters into the equation

    0 = U^2 - 2 * 9.81 * 60

    U^2 = 1177.2

    U = 34.3 m/s

    The initial velocity of the second stone is 34.3 m/s
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