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26 December, 14:27

A bolt is dropped from a bridge under construction, falling 94 m to the valley below the bridge. (a) how much time does it take to pass through the last 26 % of its fall? what is its speed (b) when it begins that last 26 % of its fall and (c) just before it reaches the ground?

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  1. 26 December, 14:30
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    Refer to the diagram shown below.

    When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.

    The last 26% of its fall is at a height of 0.26*94 = 24.4 m.

    At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.

    The time, t, for the bolt to fall a known distance obeys the equation

    s = Vt + (1/2) gt²,

    where

    s = 69.56 m, vertical distance traveled, and

    g = acceleration due to gravity.

    Therefore

    69.56 = 0 + (1/2) * 9.8*t²

    t² = (69.56*2) / 9.8

    t = 3.7677 s

    The total time, T, to fall 94 m is given by

    94 = (1/2) * 9.8*T^2

    T² = 19.1837

    T = 4.38 s

    The time taken to pass through the last 26% of its fall is

    T - t = 4.38 - 3.7677 = 0.6122 s

    The speed after falling 69.56 m is given by

    V₁ = 0 + g*t = 36.9235 m/s

    The speed with which the bolt strikes the ground is given by

    V₂ = 0 + g*T = 9.8*4.38 = 42.924 m/s

    Answer:

    (a) The bolt takes 0.6122 s to pass through the last 26% of its fall.

    (b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).

    (c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).
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