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18 July, 06:16

A falling stone is at a certain instant 90 feet above the ground. two seconds later it is only 10 feet above the ground. if it was thrown down with an initial speed of 4 feet per second, from what height was it thrown?

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  1. 18 July, 07:32
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    The equation relevant to this is:

    S (t) = So + Vot - At²/2

    Therefore we can create two equations:

    S (t) = 90 = So - 4t - 16.1t² - - > eqtn 1

    S (t+2) = 10 = So - 4 (t+2) - 16.1 (t+2) ² - - > eqtn 2

    Expanding eqtn 2:

    10 = So - 4t - 8 - 16.1 (t² + 4t + 4)

    10 = So - 4t - 8 - 16.1t² - 64.4t - 64.4

    10 + 8 + 64.4 = So - 68.4t - 16.1t²

    82.4 = So - 68.4t - 16.1t² - - > eqtn 3

    Subtracting eqtn 1 by eqtn 3:

    90 = So - 4t - 16.1t²

    82.4 = So - 68.4t - 16.1t²

    => 7.6 = 64.4t

    t = 0.118 s

    Therefore calculating for initial height So:

    82.4 = So - 68.4 (0.118) - 16.1 (0.118) ²

    So = 90.7 ft
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