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31 December, 06:22

A prankster drops a water balloon from the top of a building on an unsuspecting person on the sidewalk below. If the balloon is traveling at when it strikes a person's head (above the ground), how tall is the building

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  1. 31 December, 08:17
    0
    H = 28.5m

    Explanation:

    v = 23.7 m/s

    u = 0

    distance travelled use the relation,

    v^2 = u^2 + 2gs

    s = v^2/2g = 23.7^2/2x9.8 = 27 m

    The height of the building is,

    H = s + 1.5 m = 27 + 1.5 = 28.5 m
  2. 31 December, 08:48
    0
    Given:

    Initial velocity, vo = 0 m/s (at rest)

    Final velocity, vf = 22.5 m/s

    Height of the person, h = 1.5 m

    Using equations of motion,

    vf^2 = vo^2 + 2aS

    Where,

    S = distance of the motion

    22.5^2 = 0 + (2 * 9.81 * S)

    S = 506.25/19.62

    = 25.80 m

    Height of the building, H = distance of motion + height of the person

    = 25.8 + 1.5

    = 27.3 m
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