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What time are two thrown balls at the same height?

  1. Aug 29, 2014 #1
    1. The problem statement, all variables and given/known data

    A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.8 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 are the two balls in the air at the same height?

    2. Relevant equations

    [tex] x = x_o + v_ot + (1/2)at^2 [/tex]

    3. The attempt at a solution

    I tried setting two different versions of this equation equal to each other; one with information for the blue ball and one with information for the red ball, and then solve for time. I believe that I'm on the right track but that I'm somehow not properly figuring out the time offset for the blue ball. Any suggestions?
  2. jcsd
  3. Aug 29, 2014 #2

    Doc Al

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    Staff: Mentor

    Yes, you're on the right track. Let t be the time as measured from when the red ball was thrown. In terms of t, what is the time from when the blue ball is thrown?
  4. Aug 29, 2014 #3
    Would it be [tex] t + 0.6 [/tex] ?
  5. Aug 29, 2014 #4


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    If t is the time after the red ball is thrown and T is the time after the blue ball is thrown, then can you find the relationship between t and T?
  6. Aug 29, 2014 #5

    Doc Al

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    Staff: Mentor

    No. Note that the second ball is not thrown until 0.6 seconds after the first. So, for example, if 10 seconds has passed since the first ball was thrown (thus t = 10), how long ago was the second ball thrown?
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