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Ap physics b question (high school)

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  1. Sep 25, 2011 #1
    a stone is dropped from a 75m building. when this stone has dropped 15m, a second stone is thrown downaward with an initial velocity such that the two stones hit the ground at the same time. what was the initial velocity of the second stone?

    please use kinematic equations. that is all i know on this subject. thanks in advance.
     
  2. jcsd
  3. Sep 25, 2011 #2

    Doc Al

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    Why don't you give it a try? Use kinematic equations, of course.

    Start by figuring out how long it takes for the first stone to reach the ground.
     
  4. Sep 25, 2011 #3
    well the first stones velocity is 38.340m/s and the time is 3.912s. but idk where to go from there..
     
  5. Sep 25, 2011 #4

    Doc Al

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    OK. How long does it take for the first stone to reach 15 m? Then figure out how much time the second stone has to make it to the ground.
     
  6. Sep 25, 2011 #5
    i am confused on something. i didntt understand the problem well enough so i restarted my work. for the 1st stone, is the final velocity 0? if so, then when i used the v=v(initial)-at, my equation was 0=0+(-9.8)(t) and i got t=0. that is obviously false but idk what i did wrong.. i am terribly confused ):
     
  7. Sep 25, 2011 #6

    Doc Al

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    No, but its initial velocity is zero, since it is just dropped, not thrown.
     
  8. Sep 25, 2011 #7
    but when the stone is at 0m, isnt its velocity 0?
     
  9. Sep 25, 2011 #8

    Doc Al

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    No. It just fell 75 m, so why would you think its velocity would be zero? (We're talking about just before it hits the ground.)
     
  10. Sep 25, 2011 #9
    oh right. so i got t=2.161 for the2nd stone to hit the ground. i used thex=x(not)=v(not)^2+(1/2)at^2 and got v(not)=-28.654. but according to my answers sheet its supposed to be -24.34. i have no idea what i did wrong..
     
  11. Sep 25, 2011 #10

    Doc Al

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    OK.
    That should be:
    [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]
     
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