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22 December, 22:15

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return

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  1. 22 December, 22:43
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    36s

    Explanation:

    Let the objects be A and B.

    Let the initial velocity of A be U and the initial velocity of B be 3U

    The height sustain by A will be;

    The final velocity would be zero

    V2 = U2-2gH

    Hence

    0^2 = U2 - 2gH

    H = U^2/2g

    Similarly for object B, the height sustain is;

    V2 = (3U) ^2-2gH

    Hence

    0^2 = 3U^2 - 2gH

    U2-2gH

    Hence

    0^2 = U2 - 2gH

    H = 3U^2/2g

    By comparism. The object with higher velocity sustains more height and so should fall longer than object A.

    Now object A would take;

    From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;

    V=10*12=120m/s let g be 10m/S2

    Similarly for object B,

    The final velocity for B when it's falling it should be 3*that of A

    Meaning

    3V = gt

    t = 3V/g = 3 * 120/10 = 36s
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