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Catch the Ball

  1. Sep 12, 2006 #1
    A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.

    How fast will the person have to run to catch the ball just before it hits the ground?


    Ok, I have used the quadratic formula to come up with my Time (6.99 seconds) and then divided it into 70m, and came up with 10.00m/s for the V of the person. This is incorrect per my physics book.

    I'm thinking the problem occured on the -x part of this equation:

    [​IMG]

    I put 0, since the ball didn't move anywhere - all it did was go up and down. Any ideas where my source of error occured at? Thank you so much!
     
  2. jcsd
  3. Sep 12, 2006 #2

    Hootenanny

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    I am assuming you started with the equation [itex]s = ut + \frac{1}{2}at^{2}[/itex], I solved this equation and found t to be equal to 6.12 seconds. You may want to check your arithmetic. Your method looks good to me. Nice presentation btw :smile:
     
  4. Sep 12, 2006 #3
    hmm, you are correct. What is 'u' supposed to represent again? I thought I had these formulas down - apparently not.
     
  5. Sep 13, 2006 #4

    Hootenanny

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    u in my equation is equal to your v0 or initial velocity. Apologies for the confusion I should have stuck to your convention.
     
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