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Speed of a Baseball

  1. Jun 21, 2006 #1
    Hi all.

    I have a problem that states -

    It's the bottom of the ninth with two outs and the winning runs on base. You hit a knee-high fastball that just clears the leaping third baseman's glove. He is standing 24 m from you and his glove reaches to 3.1 m above the ground. The flight time to that point is 0.61 s. Assume the ball's initial height was 0.60 m. Find the initial speed of the ball.

    Find the time at which the ball reaches its maximum height.

    I found the initial velocity to be 39.67 m/s and this was correct (This homework is online so I can enter in answers and get results immediately.)

    I then found the angle of the launch by taking the inverse tangent of the height, 3.1 m, divided by the length, 24 m. This should give me the angle of the launch, but I'm thinking this might be where I went wrong. I then took the sine of this angle and multiplied it by the initial velocity to get my velocity in the y direction.

    Then I used the formula -

    V(final) = V(inital) + A * t

    and plugged in zero for the final velocity, since the ball's velocity in the y direction at the peak of the arc will be zero, the y velocity I calculated above for the V(initial), -9.81 m/s^2 for the acceleration due to gravity, and solved for the time. This came out to be 0.518 seconds, which was not the right answer.

    I think the part I am messing up is the calculation of the angle, but I'm not sure...any help would be greatly appreciated. :smile:
  2. jcsd
  3. Jun 21, 2006 #2


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    Staff Emeritus
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    HINT: Try finding the horizontal component of your velocity. (v = ds/dt)
    Last edited: Jun 21, 2006
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