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Phyiscs 211, two balls thrown

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A red ball is thrown down with an initial speed of 1.0 m/s from a height of 27.0 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.8 m/s, from a height of 0.8 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
    How long after the red ball is thrown, will the two balls be in the air at the same height.

    2. Relevant equations
    X=X(initial)+v(initial)t+1/2at^2


    3. The attempt at a solution
    I set the equations equal to eachother for each ball (making x the same) so I got..
    27+1t+4.9t^2=24.8(t-.4)+4.9(t-.4)^2
    After factoring and so on, I got to..
    35.96=19.88t
    t=1.8
    This is not correct, where am I going wrong?
     
  2. jcsd
  3. Jan 13, 2013 #2

    cepheid

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    Welcome to PF n_nason!


    You're not being consistent with signs. Choose upward to be either the positive x-direction, or the negative x direction, but stick with your choice.
     
  4. Jan 13, 2013 #3
    So say I make the second half of my equation...
    -24.8(t-.4)-4.9(t-.4)^2
    Then my equation would be...
    1t+4.9t^2+27=-24.8(t-.4)-4.9(t-.4)^2
    My acceleration would not cancel, is that the right track?
     
  5. Jan 13, 2013 #4

    cepheid

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    What happened to the 0.8 metres?

    EDIT: also, the acceleration is in the SAME direction for both balls (downward). So why do the two 4.9s have different signs? Gravity didn't suddenly become upward.

    Also, if your origin (x = 0) is at the ground level, and you've chosen upward to be the negative x-direction, then all of your positions should be negative as well.
     
  6. Jan 13, 2013 #5
    Oh, right that makes sense.
    Still not getting the correct answer,
    1t+4.9t^2+27=-24.8t+9.92+4.9t^2-3.92t+.784+.8
    is my simplified answer
    canceling acceleration and solving for t I get,
    29.72t=-15.496
    t=-1.92
     
  7. Jan 13, 2013 #6

    cepheid

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    Again, you chose upward to be the negative x-direction, so all of your positions should be negative.
     
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