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Sound waves of stone in a well

  1. Apr 22, 2004 #1
    Hello I am trying to solve a problem:

    You drop a stone into a well and hear the splash 1.47 s later. How deep is the well?

    Answer is 10.2 but how do I solve it?

    D = (0.5) gt squared

    D = VT
     
  2. jcsd
  3. Apr 22, 2004 #2
    Let d be the distance the particle falls. Let T be the time take for the ball to drop and the sound to reach the observer. Let t1 be the time taken for the particle to fall, and let t2 be the time taken for the sound to reach the top of the well. Also let v be the speed of sound.

    [tex]\[
    T = t_1 + t_2
    \]
    [/tex]

    [tex]\[
    d = \frac{1}{2}gt_1^2 = vt_2
    \]
    [/tex]

    [tex]\[
    \frac{1}{2}gt_1^2 = v(T - t_1 )
    \]
    [/tex]

    [tex]\[
    \frac{1}{2}gt_1^2 + vt_1 - vT = 0
    \]
    [/tex]

    [tex]\[
    t_1 = \frac{{ - v \pm \sqrt {v^2 + 2gvT} }}{g}
    \]
    [/tex]

    Now take the positive root and use that value of t1 in the formula for d and you get,

    [tex]\[
    d = \frac{1}{2}g(\frac{{ - v + \sqrt {v^2 + 2gvT} }}{g})^2 = 10.2
    \]
    [/tex]
     
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