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Kinetics ahhh

  1. Sep 26, 2004 #1
    ive been having difficulty with this problem also:

    A ball is thrown downward with an initial speed of 25m/s from the top of a 210 m tall building. At the same time, another ball is thrown upward from ground level with a speed of 25m/s. At what distance from the bottom of the two balls pass each other?

    the answer is supposed to be "-10 m so it is illogical."
    so far i know that for the first ball Vi=25m/s d=210m a=9.8m/s^2

    2nd ball Vi=25n/s d=? a=-9.8m/s^2

    are we supposed to set the distance of each of the balls equal?? i really have no idea how to do this :confused:
  2. jcsd
  3. Sep 26, 2004 #2


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    Write the position as a function of time for each of the balls then set them equal to find the time at which they meet.

    I got 18.56 m.
  4. Sep 26, 2004 #3


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    Find a equation for each ball in function of distance and equal them.Look up Kinematic Equations for Uniform Acceleration, here the acceleration is Gravity (might want to look up Free Fall, too).
  5. Sep 26, 2004 #4
    so the equation would be (25m/s)(t)+1/2(9.8)t^2=(25m/s)(t)+1/2(-9.8)t^2?
    (1st ball and 2nd ball respectively)
    what do i do with the 210 m?

    btw, my teacher gave us an answer sheet and the answer is -10m
    Last edited: Sep 26, 2004
  6. Sep 26, 2004 #5


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    Saiyan..... Stick to your Sign Convention... if put down and left is negative and up and right is positive, then this should affect the systems you're studying, i see a positive acceleration for one and a negative for another.
  7. Sep 26, 2004 #6


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    I also noticed improper use of the equation.

    [tex] Y - Y_{o} = V_{o}t + \frac{1}{2}at^2 [/tex]

    where, Yo, Vo and a is known.
  8. Sep 26, 2004 #7
    that is the equation i used: d=Vi(t)+1/2a(t)^2
  9. Sep 26, 2004 #8


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    For this problem you'll have to use it with Y - Yo.
  10. Sep 26, 2004 #9
    Y-Yo? i only see one y and thats 210m
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