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Kinematics involving two balls travelling at opposing directions

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A blue ball is thrown upward with an initial speed of 20.4 m/s, from a height of 0.5 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.8 m/s from a height of 23.9 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 blue ball is thrown are the two balls in the air at the same height


    2. Relevant equations
    Xf=Xi+Vit+0.5a^2


    3. The attempt at a solution
    Blue ball:
    Vi=20.4 m/s
    a=-9.81
    Xi=0
    t=t-2.5

    Red ball:
    Vi=-9.8
    a=-9.81
    xi=23.4
    t=t

    The height is different because i chose the blue ball as reference point.
    I plugged the values into the equation and equated them giving me:
    20.4(t-2.5)-4.9(t+2.5)^2=23.4-9.8t-4.9t^2

    further simplifying i get....
    -4.1t-4.9t^2-81.6=23.4-9.8t-4.9t^2
    then...
    5.7t=105
    t=18.4s

    which is wrong. can someone tell me where i went wrong?
     
    Last edited: Sep 9, 2012
  2. jcsd
  3. Sep 9, 2012 #2

    TSny

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    Does the blue ball travel for more time or less time than the red ball?
     
  4. Sep 9, 2012 #3

    gneill

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    Staff: Mentor

    Okay, those are the facts, but what is the question? What are you trying to determine?
     
  5. Sep 9, 2012 #4
    The ball traveled for a longer time so i switched the signs but i still got the wrong answer.
     
  6. Sep 9, 2012 #5

    TSny

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    The question asks for the time after the blue ball is thrown. Note that you defined t so that it gives the time after the red ball is thrown. Did you take that into account?
     
  7. Sep 10, 2012 #6
    okay i got the answer but i'm not sure why i had to add 2.5 second to the "t" that i found. Is it because the t was referring to the time that the red ball took, but the blue ball got a 2.5s head start?
     
  8. Sep 10, 2012 #7

    TSny

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    Yes, that's right. If you look at your original post you can see where you let t stand for the time the red ball traveled and (after correction) you let t + 2.5 represent the time for the blue ball.
     
  9. Sep 10, 2012 #8
    if i let the red ball's t be t-2.5 will it give me the same answer?
     
  10. Sep 11, 2012 #9
    It is easier to see if you make the reference point as the position and velocity of after 2.5s since the red ball exists only on 2.5th sec.

    As for your calculation the time for blue ball should be (t) and time for red ball is (t-25)
     
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