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If a rock is dropped off a sea cliff

  1. Oct 6, 2003 #1
    If a rock is dropped off a sea cliff, and the sound of it striking the water is heard 3.4 seconds later, how high is the cliff, assuming the speed of sound to be 340 m/s.
    I don't really know where to begin with this one, and I would really appreciate the help. Thanks.
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Oct 6, 2003 #2


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    3 equations in 3 unknowns. You do the algebra.

    1) h=gs2/2
    2) h=vt
    3) s+t=3.4

    h=height of cliff
    g=gravitational constant (9.83 m/s2)
    s=time for rock to fall
    t=time for sound to come back to you
    v=speed of sound (340 m/s)

    (h,s,t) are unknowns.
  4. Oct 6, 2003 #3
    Thanks for the help. As I said before, my difficulty lay in not being able to start the problem, and now I know where I must begin. Thanks again for helping me out.
  5. Oct 18, 2003 #4
    It's intersting that your value for g is 9.83. Was that a typo or is that what you've been given. I've always been told that it was 9.81.
  6. Oct 19, 2003 #5
    You live in England, so the gravitational acceleration there is 9.81 m s-2. In other places the value of g is different, due to the fact that Earth is not a perfect sphere.

    Where I live, in Singapore which is near the equator, the value of g is 9.78 m s-2, although in classrooms we use 9.81 m s-2 because the public education here prepares us for Cambridge papers.
  7. Oct 19, 2003 #6
    I know that the earth is not a perfect sphere, and that invariably gravity will vary. But I thought that 9.81 was an average figure denoted by SI. After all, they set the standard units for everything else so why not a standard value for g aswell.
  8. Oct 19, 2003 #7
    Hmmm... I've never thought of that. I don't know...
  9. Oct 21, 2003 #8
    Because the other things that they set values for are constants (or are believed to be), whereas 'g' certainly isn't. If I measure 'g'in the lab, I want to find the value of it, not see how close I am to some arbitrary average decided on by a committee.
  10. Oct 23, 2003 #9


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    g is not an arbitrary unit like "1 kg" or "1 meter". It is a "constant of nature" and the correct value, to the necessary accuracy, should be used for the position on the earth. Of course, for most applications, 9.81 gives the necessary accuracy. (And, in fact, 9.8 works nicely most of the time.)
  11. Oct 24, 2003 #10


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    I hadn't realized it changed so much so I looked it up. The difference is a combination of distance from the center of the earth and centrifugal force. I don't feel like doing the math right now on exactly how much of it is due to each.
  12. Oct 24, 2003 #11
    The formulae is:

    F' = GMm

    F' is the force of attraction between two point masses of M and m.
    G is the gravitational constant (6.67*10^24).
    r is the distance between the two point masses.

    You could take account of the fact that we're trying to mave in a straihg line and the earth is pulling us round in a circle if you wanted to, but this effect would be smal when you take account of the radius of the earth: 40,000,000/(2[pi]), nevertheless the formula for that is:

    F'' = mr * ω^2

    where &omega is the angular velocity, which can be replaced by 2[pi]/T where T is the time persiod 24*3600 (24 hours in seconds), m is your mass by the way:

    F'' = mr * 4 * [pi]^2 / (24*3600)^2

    So the total resultant force on a mass m, F = F' - F''

    Feel free to work out the values if you can be bothered.
  13. Jan 25, 2004 #12
    Excuse me, but g is not a constant. It is commonly called the acceleration due to gravity, which is related to the Gravitational constant, G. The value of g depends on where it is measured, including the altitude.

    Even though g is commonly called the acceleration due to gravity, the measured (and published) value of g at a certain location usually includes the effect due to the rotation of the Earth at the particular location.
  14. Jan 25, 2004 #13
    the height is 578m.

    Δy = [(v+v0)/2]t

    v = 0, t = 3.4s.
  15. Feb 2, 2004 #14
    Hi Adrian

    Its Magg$,

    you know, we talked before and you helped me with my coursework...

    Only this time I'm talking out of my own interest...

    Could you tell me what the Young's Modulus value is for:
    and Titanium...

    I know your gonna think this is to help with coursework again because Young's Modulus is part of the AS syllabus but HONESTLY, this is through my own interest as I want more young's modulus values to do some comparisons between materials.

    Please Help,


    P.S, I'm going to a lecture tomorrow about the Physics of Skydiving at Birmingham University, should be good!
  16. Feb 2, 2004 #15


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    Here's a site with some of them. You can probably find more with a little judicious googling. I found this one just by googling on Young's Modulus. You should generally try that kind of thing before asking here, because you get the answer quicker and (hem) learn to do independent research.
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