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13 September, 05:07

A foul ball is hit straight up into the air with a speed of a bout 25 m/s (a) how high does it go? (b) how long is it in the air

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  1. 13 September, 05:11
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    The relevant equations we can use in this problem are:

    d = v0 t + 0.5 a t^2

    v^2 = v0^2 + 2 a d

    a. how high does it go?

    We are to find for the distance d. The formula is:

    v^2 = v0^2 + 2 a d

    where v is final velocity = 0 at the peak, v0 = 25 m/s, a is negative gravity since opposite to direction = - 9.8 m/s^2, d is height achieved = ?

    0 = (25) ^2 + 2 (-9.8) d

    d = 31.89 m

    Therefore the maximum height is about 31.89 meters

    b. how long is it in the air

    We use the formula:

    d = v0 t + 0.5 a t^2

    in going up, a = - g = - 9.8 m/s^2, find for time t:

    31.89 = 25 t + 0.5 ( - 9.8) t^2

    -4.9 t^2 + 25 t = 31.89

    t^2 - 5.1 t = 6.51

    Completing the square:

    (t - 2.55) ^2 = 6.51 + 6.5025

    t - 2.55 = ± 3.61

    t = - 1.06, 6.16

    Since time cannot be negative, therefore:

    t = 6.16 seconds

    However this is not the total time yet, because this only accounts for the time going up.

    The time going up and going down are equal therefore:

    t (total) = 2 t

    t (total) = 12.32 seconds

    Therefore the foul ball spent 12.32 seconds in the air.
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