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13 March, 20:47

A bolt is dropped from a bridge under construction, falling 97 m to the valley below the bridge. (a) how much time does it take to pass through the last 11 % of its fall? what is its speed (b) when it begins that last 11 % of its fall and (c) just before it reaches the ground?

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  1. 13 March, 23:09
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    The first thing we have to do for this case is write the kinematic equationsto

    vf = a * t + vo

    rf = a * (t ^ 2/2) + vo * t + ro

    Then, for the bolt we have:

    100% of your fall:

    97 = g * (t ^ 2/2)

    clearing t

    t = root (2 * ((97) / (9.8)))

    t = 4.449260429

    89% of your fall:

    0.89*97 = g * (t ^ 2/2)

    clearing t

    t = root (2 * ((0.89 * 97) / (9.8)))

    t = 4.197423894

    11% of your fall

    t = 4.449260429-4.197423894

    t = 0.252

    To know the speed when the last 11% of your fall begins, you must first know how long it took you to get there:

    86.33 = g * (t ^ 2/2)

    Determining t:

    t = root (2 * ((86.33) / (9.8))) = 4.19742389 s

    Then, your speed will be:

    vf = (9.8) * (4.19742389) = 41.135 m / s

    Speed just before reaching the ground:

    The time will be:

    t = 0.252 + 4.197423894 = 4.449423894 s

    The speed is

    vf = (9.8) * (4.449423894) = 43.603 m / s

    answer

    (a) t = 0.252 s

    (b) 41,135 m / s

    (c) 43.603 m / s
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