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Waves & Sound: How deep is your well?

  1. Nov 20, 2004 #1
    I'm working on a problem for my Grade 11 class.

    A stone is dropped down a well, and, 16 seconds later, the sound of it splashing into the water is heard by people at ground level. The temperature in the well is 10 degress celsius.

    I'm asked to find the depth of the well. So, what I've done so far is this...

    I've said the change in time for the falling rock will be 'y' and the change in time for the sound travelling back up will be '16 - y'. The distance from the top of the well to the point of impact (source of the sound) is 'x'. Given negligible air resistance and whatnot, a = g. The speed of the sound returning up the well will be 332 m/s + 0.6 (10 degrees C), or 338 m/s.

    I've taken my kinematics equations, and said that, for the falling rock, x = 1/2gy^2, and for the sound travelling up, x = 338 m/s * (16-y). I've set the two equal, but I can't do anything past this point. For some reason, I can't remember the algebraic ... uh, stuff.. that I'd need to do this. I'm thinking I should convert what I have to a quadratic (a + b -c = 0?) and then use the quadratic equation, but I can't remember how to do that.

    Thanks in advance.
  2. jcsd
  3. Nov 20, 2004 #2
    1/2*9.8*y^2 = 338(16-y) is what you have. Converting to ay^2 + by + c = 0 shouldn't be too bad: Just distribute the 338 on the right hand side and subtract it over to the left hand side.
  4. Nov 20, 2004 #3


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    Looks like you have done the hard part correctly, this getting the correct relationships. I am surprised that you can't finish the simple algebra!
    You have:
    [tex] x = \frac {gy^2} 2 = v_s ( 16-y) [/tex]
    ( vs is the speed of sound )
    so now all you need do is assemble the pieces to get the standard from of a quadratic.
    First distribute vs across the RHS (Right Hand Side)

    [tex] x = \frac {gy^2} 2 = v_s 16- v_s y [/tex]
    Now add
    [tex]x = -(v_s 16-v_s y) [/tex]

    to each side, to get,

    [tex] \frac {gy^2} 2 + v_s y - 16 v_s = 0 [/tex]

    Now for a refresher

    if you have a quadratic

    [tex] ax^2 + bx +c =0 [/tex]
    The roots are:

    [tex] x = \frac { -b \pm \sqrt{b^2 - 4 ac}} 2a[/tex]

    Can you complete your problem now?
  5. Nov 20, 2004 #4
    Mmm..awesome. Yeah, I can do this now. Thanks!
  6. Nov 21, 2004 #5
    just to be clear once you have the change in time for the falling rock you can the put that time into d= at^2/2 correct? just for future reference, what is the correct answer?
  7. Nov 21, 2004 #6
    Just to avoid any confusion, I believe you meant the a to be in the denominator as well.
    Last edited: Nov 21, 2004
  8. Nov 22, 2004 #7


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    As you well know you are correct! Unfortunately, it is now to late to fix the error so it must stay as is.
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