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4 January, 05:25

a soccer ball is kicked horizontally off a 22 meter high hill and lands 35 meters from the edge of the hill. determine the initial horizontal velocity of the soccer ball

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  1. 4 January, 09:12
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    y = y0 + v0*t + 0.5at^2

    where y0 = initial vertical position = 22m

    y = final vertical position = 0m

    v0 = initial vertical velocity = 0 m/s

    a = acceleration = - 9.8 m/s^2

    t = time in seconds

    0 = 22 + 0*t + 0.5 (-9.8) t^2

    t^2 = 22/4.9 = 4.49 s^2

    t = 2.12 s

    So it traveled 35m in 2.12 s

    the horizontal distance traveled is determined by:

    x = x0 + v0*t + 0.5at^2

    but here a in the horizontal direction is 0 m/s^2

    and v0 is in the velocity in the horizontal direction in this equation

    35 m = 0 + v0*t

    35 m = v0 (2.12 s)

    v0 = 16.5 m/s

    So the ball was kicked 16.5 m/s in the horizontal direction
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