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Homework Help: COnstant acceleration - stone in a well

  1. Sep 9, 2011 #1
    1. The problem statement, all variables and given/known data

    You drop a stone down into a well to find the depth of the well. You hear the sound of the splash 3.5 s after you let the stone go.

    2. Relevant equations

    [tex]s(displacement)=v_0t \cdot \frac 12 at^2[/tex]

    3. The attempt at a solution

    I have
    [itex]t_{total}=3.5 s[/itex]

    Stone: [itex]v_0[/itex]=0, and a=9.81 m/s^2.

    Sound: v=340 m/s, a=0.

    If the sound had not used a small part of the total time, the answer would have been:

    [tex]s(displacement)=v_0t \cdot \frac 12 at^2=0*3.5 s + \frac 12 * 9.81 m/s^2*(3.5s)^2=60 m[/tex]

    But the answer key of my book says 55 m. I see why, but I don't know how I can get that answer... Please help, anybody out there!
  2. jcsd
  3. Sep 9, 2011 #2


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    Gold Member

    basically in this question there is 3 unknowns and 3 constraints

    3 unknowns
    - time for rock to reach bottom of well
    - time for sound to travel out of the well
    - depth of well

    3 constrants
    - time takes to reach the bottom of the well and depth of the well is related to the equation you used
    - time taken for sound to travel out of the well follows a simple s = vt relationship
    - total time is 3.5s

    can you see how to solve this now?
  4. Sep 9, 2011 #3
    Hmmm, not quite sure. Depth of well, s, is the same both when the stone and the sound travels it.

    Sound: [itex]s(displacement)= vt = 340m/s *t_{sound}[/itex]


    [tex]s(displacement)=v_0t\cdot \frac 12 at^2=0∗(3.5s-t_{sound}) +\frac 12 ∗ 9.81m/s^2 ∗ (3.5s-t_{sound})^2=0+12∗9.81m/s^2∗(3.5s-t_{sound})^2=12∗9.81m/s^2∗(3.5s-t_{sound})^2[/tex].

    I can relate these two equations to each other: [itex]340m/s *t_{sound}=12∗9.81m/s^2∗(3.5s-t_{sound})^2[/itex] and solve this equation for [itex]t_{sound}[/itex]. Am I on the right track now?
  5. Sep 9, 2011 #4


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    looks like you are :)
  6. Sep 9, 2011 #5
    That was right :)

    Thanks a lot !!!
  7. Sep 9, 2011 #6


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    Gold Member

    glad to help :biggrin:
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