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Homework Help: Ball Freefall problem

  1. Sep 20, 2007 #1
    A ball is thrown vertically upward with an initial speed of 15 m/s. Then, 0.92 s later, a stone is thrown straight up ( from the same initial height as the ball) with an initial speed of 27 m/s. The acceleration of gravity is 9.8 m/s^2.
    How far above the release point will the ball and stone pass each other? Answer in units of m.

    So far, I have written down this equation 15( t + 0.92) - 4.9(t + 0.92)^2= 27t- 4.9t^2
    And when I simplify everything, I get a t= .4593

    Now, I don't know if that is even remotely correct, but if I am, I would like to know how to go about getting closer to the answer at hand.

    Thank you.
  2. jcsd
  3. Sep 20, 2007 #2


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    Sounds right, now just substitute t into one of the equations and get the height - better still do both equations just to check the answer
  4. Sep 20, 2007 #3
    Its correct.As the displacement of both the oblects are same you can use the equality among the equations as you have done above.
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