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5 February, 20:17

A ball thrown horizontall at 22.2 m/s from the roof of a building lands 36.0m from the base of the building. How tall is the building?

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  1. 5 February, 20:19
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    Velocity of the ball (v) = 22.2 m/s

    Distance from the base of the building where the ball falls (d) = 36.0m

    We know that, Velocity (v) = distance (d) / time (t)

    From the above equation,

    Time (t) = distance (d) / velocity (v)

    Substituting the values for distance and velocity in the above equation,

    Time = 36.0/22.2 = 1.62 seconds

    When the ball is falling down, it accelerates due to gravity

    This acceleration (a) = 9.8 m/s²

    Now we need to find the height of the building.

    The equation of motion for an object with constant acceleration is,

    s = ut + 1/2 at²

    Where,

    s = vertical displacement, which is height of the building in our case

    u = initial velocity of the ball = 0

    Substituting all the above values in the equation of motion,

    s = (0 * 1.62) + (1/2 * 9.8 * 1.62²)

    s = 0 + (25.72/2) = 12.86 m = 12.9 m which is the height of the building

    Therefore the building is 12.9m tall
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