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Ball and stone thrown upward - kinematics

  1. Sep 8, 2008 #1
    Here is another question.I try to get the answer but .....

    1.A ball is trown vertically upward with an initial speed of 15m/s.then,2.2 s later ,A stone is thrown straight up (from the same initial height as the ball)with an initial speed of 21.3m/s.
    the acceleration of g=9.8m/s^2. How far above the release point will the ball and stone pass each other?

    Here is what I did.

    I use the same time for both (2.2second). the equation is Y=Y(inital) + v(inital)t + 1/2 gt^2

    After i got both distance I subtracted. which is gave me the didtance b/n them.
     
  2. jcsd
  3. Sep 9, 2008 #2

    tiny-tim

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    solution

    Hi david12! :smile:
    ah … that's what you've done wrong.

    t is unknown :wink:

    you have to work out two equations relating y and t … one for the ball and one for the stone …

    and it helps to use different letters for them

    say y1 = f(t1) and y2 = f(t2)

    and then you solve for y1 = y2 and t1 = t2 + 2.2 :smile:
     
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