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23 June, 16:48

A 1.0-kilogram ball is dropped from the roof of a building 40. meters tall. What is the approximate time of fall? [Neglect air resistance.]

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  1. 23 June, 19:32
    0
    H = 40 m, the height from which the ball is dropped.

    m = 1 kg, the mass of the ball

    Assume g = 9.8 m/s² and neglect air resistance.

    The initial vertical velocity is zero.

    If t = the time of flight, then

    40 m = (1/2) * g * (t s) ² = 0.5*9.8*t²

    t² = 40/4.9 = 8.1633

    t = 2.857 s

    Answer: 2.9 s (nearest tenth)
  2. 23 June, 20:45
    0
    2.856s Distance traveled under constant acceleration as a function of time is given by x = (gt^2) / 2 where g is the acceleration and t is time. In this case acceleration is due to gravity and is 9.81m/s^2. The distance of interest is x=40m. Substituting and solving 40 = (9.81*t^2) / 2 80 = 9.81*t^2 8.1549 = t^2 2.856s = t
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