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a baseball is popped straight up. the ball leaves the bat moving at 37.8 m/s. to what height did the ball rise.

t=7.72 s

the answer is 73 m, but how do you get that answer

  • Physics -

    initial KEnergy=final PEnergy
    1/2 m v^2=mgh

    h= 1/2 *v^2/g=1/2 * 37.8^2/9.8 in my head about = 1600/20=80 meters, work it out accurately. 73 meters looks good

  • Physics -

    You can use the formula for Uniformly Accelerated Motion (UAM):
    h = (v,o)t - (1/2)gt^2

    Or you can use another formula, which does not use time:
    v,f^2 - v,o^2 = 2gd
    v,f = final velocity
    v,o = initial velocity
    g = acceleration due to gravity = 9.8 m/s^2
    d = distance (in this case, it's the height)

    Note that at the highest point, the ball does not move (v,f = 0). We're solving for d. Substituting to the second equation:
    0 - (37.8)^2 = 2(-9.8)(d)
    -1428.84 = -19.6*d
    d = 1428.84 / 19.6
    d = 72.9 m

    The negative sign of g indicates the direction (since acceleration is a vector quantity).
    Hope this helps :)

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