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Physics Kinematics of a rock drop

  1. Sep 13, 2005 #1
    Hey... need some help with this quesiton:

    A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.2 seconds later. Ifthe speed of sound is 340 m/s, how high is the cliff?

    multiplying 3.2 by 340 gets me 1088 m which doesnt seem right.

    How do I go about solving this problem?

    Thanks much.
  2. jcsd
  3. Sep 13, 2005 #2
    That does not seem right. The question seems rather ambiguous as to whether they mean 3.2 seconds after the rock strikes the ocean or after it is dropped. Try it the other way; 3.2 seconds after the rock is dropped
    Last edited: Sep 13, 2005
  4. Sep 13, 2005 #3
    There was already a thread on this. Try searching.
  5. Sep 13, 2005 #4
    I read the other thread but I still dont get it.

    H(t) = -4.9t^2+ h0

    I do not know h0 so that remains an unknown. IF I add that to v/h then Im gonna have quite a few unknowns.
  6. Sep 13, 2005 #5


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    Is h0 the initial speed? If so, shouldn't it be zero?
  7. Sep 13, 2005 #6
    H0 is imo the initial height.

    If h0 = 0 ,

    H= (-4.9)(3.2) = 15.68 m

    H+ (v/h) = 15.68 + ( 340/15.68) = 37.36 m

    Is that correct?
  8. Sep 13, 2005 #7
    haha are you in my class?

    number 47 right Maccully
  9. Sep 13, 2005 #8
    huh...........? whats 47? This is from the Giancoli book by Prentice Hall.
  10. Sep 13, 2005 #9
    its number 47, in the book
  11. Sep 13, 2005 #10
    probably is.. dunno.... anyhow, still need the answer.. got a test tomorrow on this chapter :(
  12. Sep 13, 2005 #11
    Time spent falling + time for sound to travel up cliff = 3.2s

    To work out the time spent falling reaarange your equation of motion
    The time for the sound to travel up the cliff is x/v, so your equation is:
    [tex]\sqrt{\frac{2x}{a}} + \frac{x}{v} = 3.2[/tex]
    which has a solution at 46.02m
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