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Two Dimensional Motion (I)

  1. Feb 15, 2007 #1
    I am stumped big time!

    Two runners start at the same point on a straight track. The first runs with a constant acceleration so that he runs 93 yards in 8 seconds. The second runner waits 3 seconds and then throws a rock at his opponent's head. If the head and the rock are at the same level form the grownd, what must the inital magnitude of the velocity be if the rock is to hit the head just at the 93 yard tape?
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  3. Feb 15, 2007 #2


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    What distance does the rock need to cover in how much time?
  4. Feb 15, 2007 #3
    The same distance as the runner who actually runs covers but the time is unknown
  5. Feb 15, 2007 #4


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    Is it? The stone must hit the runner as he reaches the finish line. You know the time that this happens, so can you say what the time that the stones takes in the air is?
  6. Feb 16, 2007 #5
    the stone takes 5 secs in the air
  7. Feb 16, 2007 #6


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    Yup, so now can you set up a relevant equation?
  8. Feb 16, 2007 #7
    how do you get the vertical component of velcocity.

    you get the horizontal component of velocity with what you have - 93/5 = 18.6 yrds/sec
  9. Feb 16, 2007 #8


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    For the horizontal component, you have used the equation d=ut+at2/2, where u is the initial velocity. Set up a similar equation in the y direction; you know d, a and t, so you can solve for uy.
  10. Feb 16, 2007 #9
    the vertical height is not given. unless we use zero
    then we get 24.5
  11. Feb 16, 2007 #10
    Please can you explain this because it makes no sense. We don't know the vertical distance
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