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This simple problem gives me fits

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A rock falls of the cliff, 3.4 seconds later, you can hear the rock hitting the ground. Assuming the speed of sound is 340 m/s, how high is the cliff?


    2. Relevant equations
    Delta X = Vot + 0.5at^2
    2a Delta X = V^2 - Vo^2


    3. The attempt at a solution
    I don't know how to start. The height of the cliff is unknown and the time it took the rock to hit the ground is unknown. Initial velocity is zero and that's all I can say. I don't know where to start. Perhaps I am missing something.
     
  2. jcsd
  3. Sep 21, 2010 #2
    The first thing you need to do is quantify your total time 't' as the sum of the time it takes for the rock to fall 'tf' plus the time it takes for the sound to travel back 'ts' or t=tf+ts. Then you need to use your position equation h(t) where h(tf)=...
     
  4. Sep 21, 2010 #3
    Ok so tf + ts = 3.4 s

    Position equation of the tf would be [tex]\Delta[/tex]y = Vo tf + 1/2 atf^2

    Position equation of ts would be [tex]\Delta[/tex]y = v ts (where v = 340 m/s)

    Correct?
     
  5. Sep 21, 2010 #4
    Yes, that is pretty much correct. The only thing is your delta y is really height as a function of time or y(t)=vo*t+1/2a*t^2 so when you sub in your total fall time 'tf' you are implying that your height is zero. To get the 'tf' term to sub into that function use the relation t=tf +ts or tf=t-ts. Now you just need your relation of the height of the cliff and the velocity of sound to sub in for 'ts'.
     
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