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28 September, 08:25

A soccer ball is kicked with a speed of 22m/s at an angle of 35.0º above the horizontal. If the ball lands at the same level from which it was kicked, how long was it in the air?

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  1. 28 September, 08:46
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    2.57 seconds

    Explanation:

    The motion of the ball on the two axis is;

    x (t) = Vo Cos θt

    y (t) = h + Vo sin θt - 1/2gt²

    Where; h is the initial height from which the ball was thrown.

    Vo is the initial speed of the ball, 22 m/s, θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²

    We want to find the time t at which y (t) = h

    Therefore;

    y (t) = h + Vo sin θt - 1/2gt²

    Whose solutions are, t = 0, at the beginning of the motion, and

    t = 2 Vo sinθ/g

    = (2 * 22 * sin 35°) / 9.81

    = 2.57 seconds
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