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24 December, 04:46

As he is catching a 2.5 kg package, your friend exerts a downward force of 1.0x10^1 N. If the package was still moving upwards at 6.7 m/s at the start of his catch, how much time will it take to bring him to a stop?

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  1. 24 December, 05:56
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    F = ma

    A = F / m = 1 x 10^1 + (2.5) (9.3) / 2.5 = 1 x 10^1 + 24.5 / 2.5 = 34.5 / 2.5 = 13.8 m/s^2

    By using kinematic equation, V = 4 + ak

    0 = 6.7 - 13.8 t

    13.8t = 6.7

    T = 6.7 / 13.8 = 0.4855 sec
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