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Kinematics simultaneous motion problem!

  1. Sep 26, 2006 #1
    the top question reads

    "You are on top of a building 186 meters high you throw a ball downard at a speed of 18m/sec"

    After calculating time for hitting the ground final velocity and average velocity the next question says:

    "Suppose instead that at the same moment that this ball was thrown downward from the roof a second ball was thrown upward from the ground with a speed of 32m/s"

    I know that when theres simultaneous motion you are supposed to find what the two objects have in common but I can't seem to find it.

    I found the time to hit the ground for the first ball was 4.59, its final velocity was -62.98m/s and its average velocity was -40.52m/s.

    I figured the time it takes the ball from the ground to reach a velocity of 0 is 3.27seconds and the height it reaches is 52.32 meters I'm not sure what to do now though, somebody please give me a hint or steer me in the right direction. Thank you.
     
  2. jcsd
  3. Sep 26, 2006 #2
    for the ball thrown downward ...
    h = 186 - 18t - 4.9t^2

    for the ball thrown upward ...
    h = 32t - 4.9t^2

    they meet when h = h ...

    186 - 18t - 4.9t^2 = 32t - 4.9t^2

    solve for t (very easy), then find h.
     
  4. Sep 26, 2006 #3
    thanks..?

    Okay, I understand that they both will hit when h = h, but where are you getting those two equations from, and I calculated a time of 3.72 seconds for them to collide, and a height of .85 meters please let me know if this is right.



    Can anyone let me know if these answers are correct?
     
    Last edited: Sep 26, 2006
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