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Height of cliff above water

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data

    You throw a rock off a cliff, giving it a velocity of 8.3 m/s straight down. At the instant you released the rock, your hiking buddy started a stopwatch. The rock hit the water below exactly 6.9 seconds after you threw the rock. How high is the cliff above the water?

    2. Relevant equations

    Δd = 1(v1+v2)Δt
    2aΔd = v2^2 - v1^2

    Δd = v1 Δt + 1/2 a (Δt)2

    Δd = v2 Δt - 1/2 a (Δt)2

    3. The attempt at a solution

    I am assuming that the v1 is 0 because it starts from rest and v2 is -8.3m/s. (Sign convention was down negative.) and time was 6.9. I subbed everything into the equation. When I make everything equal to Δd, does that give me the height of the cliff above the river?
  2. jcsd
  3. Oct 7, 2007 #2


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    All you need is the s = ut + 1/2at^2
    But the initial velocity u=8.3 and a=g=9.8m/s^2
    Don't try and remember the sign convention, think about would S get bigger or smaller with time, would a large U reduce/increase S.
  4. Oct 7, 2007 #3
    So, it would be:

    D = (8.3m/s)(6.9s) + 1/2 (9.80/ms^2)(6.9s)^2

    Work it out and the D would give me the height?
  5. Oct 7, 2007 #4


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    Yes (and some more unnecessary text to make the anwer longer than 4 characters otherwise the forum software rejects it )
  6. Oct 7, 2007 #5
    Thanks alot this si a great help. Do you think you can help me with the other question I have posted?
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