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29 September, 14:24

From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engineering building is 35.0 m tall. what is the initial velocity of the ball?

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Answers (2)
  1. 29 September, 14:34
    0
    Equations of the vertical launch:

    Vf = Vo - gt

    y = yo + Vo*t - gt^2 / 2

    Here yo = 35.0m

    Vo is unknown

    y final = 0

    t = 4.00 s

    and I will approximate g to 10m/s^2

    => 0 = 35.0 + Vo * 4 - 5 * (4.00) ^2 = > Vo = [-35 + 5*16] / 4 = - 45 / 4 = - 11.25 m/s

    The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.

    Answer: 11.25 m/s
  2. 29 September, 14:48
    0
    The diagram shown below illustrates the problem.

    v = vertical upward launch velocity.

    g = 9.8 m/s² is acceleration due to gravity.

    When s = - 35m, the elapsed time is 4 s.

    Therefore

    ( - 35 m) = (v m/s) * (4 s) - (1/2) * (9.8 m/s²) * (4 s) ²

    -35 = 4v - 78.4

    4v = 78.4 - 35 = 43.4

    v = 10.85 m/s

    Answer: 10.85 m/s
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