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Basic physics prob.

  1. Jan 22, 2007 #1
    A stone is thrown vertically upward at a speed of 49.80 m/s at time t=0. A second stone is thrown upward with the same speed 2.670 seconds later. At what time are the two stones at the same height?

    i found that to be 6.411s

    at what hight do the two stones pass eachother?
    ok, so that hight will be at the time that they are at the same hight which is t= 6.411s

    so i used the equation deltax=Vot+1/2at^2 and i got 1023 m but that isn't right.... can someone help me out please


    what is the downward speed of the first stone as they pas eachother?
    so that time again would be 6.411

    and i used the equation v=vo+at .....so v=49.80-9.8(6.411) which is got to be -13.32 m/s but that isn't right either

    please help me
  2. jcsd
  3. Jan 22, 2007 #2
    This is just a matter of setting the equations (one of the three, you figure that out) equal to each other =).

    Y initials are the same and we want the Y finals to be the same =).

    For the third part; you should know the delta y. Then you call the very peak of it's throw where v = 0. You can now call that a new point and say Yo = 0, and the final Y is your delta Y.
    Last edited: Jan 22, 2007
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