A major-league pitcher can throw a ball in excess of 39.1 m/s. If a ball is thrown horizontally at this spe?

A major-league pitcher can throw a ball in excess of 39.1 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

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  • 9 years ago
    Favorite Answer

    Dx = horizontal distance traveled = 17.0 meters

    Vx = horizontal speed = 39.1 m/sec = 87.46 mi/h

       Time of Flight = t = (Dx) ⁄ (Vx) = (17.0) ⁄ (39.1) = 0.43478 sec

       h = height relative to initial height

     Vyi = initial vertical speed = 0 ... (ball thrown horizontally)

      h = (½) • g • t² + (Vyi) • t

      h = (½) • (-9.81) • (0.43478)²

      h = -  0.927 meters  ...  almost a meter drop-off

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    4 years ago

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