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Baseball related question

  1. Jan 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A major league pitcher can throw a baseball in excess of 42.4 m/s. if a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 15.3 m away from the point of release?


    2. Relevant equations

    v=d/t and y=1/2(v+v0)t

    3. The attempt at a solution

    time=d/t y=1/2(V+v0)t
    =15.3/42.4 =1/2(42.4+0)(.361)
    =0.361s y=7.65

    Answer: 7.65m
    Am I right?
    help me If I am wrong
    Thank You
     
  2. jcsd
  3. Jan 22, 2007 #2
    Does your answer sound reasonable - the ball drops 7.65 meters?
    You're missing the necessary equation; one that would involve acceleration (or gravity to be more specific.)
    The first part of your solution: that time is 0.361 seconds is important, although you typed one of the letters wrong in the equation (you have time = distance divided by time) Gravity will be acting on the ball for that amount of time. For what it's worth, at least in physics in NY, you would automatically lose half the credit for not substituting with the units; i.e. you have t=d/v = 15.3meters / (42.4 m/s). But, it's up to your teacher as to what's allowed.

    Incidentally, the second equation you used implies that either the ball speeds up uniformly to 42.4 m/s over the .361 second span of time while it travels from the mound to home plate, or that it slows down uniformly to 0 as it travels.
     
  4. Jan 22, 2007 #3
    I still don't get it. help me further.
    I will appreciate it.
     
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