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Two stones off a cliff

  1. Mar 13, 2007 #1
    2 stones are dropped off a 60 m cliff the second stone is droped 1.6 s after the first one. how far below the top of the cliff is the second stone when th seperation between the 2 stones is 36 m?

    so i am trying to use the equation x1=xo+tvo+.5at^2

    then x2=xo+(t+1.6)vo+.5a(t+1.6)^2

    or should i set my zero point somewhere else like where stone 2 is when the distance between the stones is 36 m?

    please give me a lead
     
  2. jcsd
  3. Mar 13, 2007 #2

    hage567

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    You need to get an expression for the height of each stone. The expression you have there is correct (if it is for the first stone), you can simplify it quite a bit though. So what would the equation of the other stone be? I would take the top of the cliff as the zero point.

    You are given the difference in height between them at a certain time, so you need to find the time t that they are 36m apart.
     
  4. Mar 13, 2007 #3
    interesting wrinkle. Offhand I would think there are a few ways to the right answer. And surely you are on the right track.

    I usually try to find the simplest physically intuitive approach. The way then i look at this problem is to "ignore" acceleration snce its the same for both and therefore cancels once both are in free fall and gaining velocity at the same rate. This does not mean that the distance between them is constant. So off the top of my head, I'd first figure the velocity after 1.6 sec=at. this is 15.68m/s. To reach 36m would require the time at this speed. Then backtrack for position. I may be all wet, is so someone willbe sure to say so.
     
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