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Velocity/Accel question

  1. Sep 10, 2008 #1
    A stone is thrown vertically upward at a speed of 25.30 m/s at time t=0. A second stone is thrown upward with the same speed 1.860 seconds later. At what time are the two stones at the same height?

    what are the equations needed, and any help is greatly appreciated thanks
  2. jcsd
  3. Sep 11, 2008 #2


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    Hi blueskadoo,

    What have you tried so far?
  4. Sep 11, 2008 #3
    i have tried setting them equal.
    also, tried using the equation X-Xo= Vox*t + .5Ax*t^2
    and Vx^2=Vox^2 + 2 Ax (x-xo)

    i cant seem to be getting anything reasonable...
  5. Sep 11, 2008 #4


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    How long does it take to reach max height?

    Would this be given by Vo/g = t seconds?

    Figure that value to decide where the first stone will be when the second is released. (If it has already hit the ground by the time you throw, then it will have already crossed before you threw it. If the first stone is dropping when you throw #2 ... it's good to know what the two are doing so you can keep things straight.
  6. Sep 11, 2008 #5
    Write out the time dependence of displacement for both the bodies (keep the 1.860 seconds in mind) and equate them....
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