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Homework Help: Kinematics/speed of sound question

  1. Jul 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A stone is dropped in a well. The splash is heard 3s later. What is the depth of the well?

    2. Relevant equations
    constant acceleration equations
    take speed of sound to be 343 m/s.

    3. The attempt at a solution
    For the stone:
    u = 0
    a = g
    s = ut + 0.5at^2

    subbing it all in you get t_stone = +sqrt(2s/g)

    For the sound wave:
    u = v = 343 m/s
    v = s/t
    subbing it all in you get
    t_wave = s/343

    (same distance travelled for stone and wave = s)

    so you have t_stone + t_wave = 3

    +sqrt(2s/g) + s/343 = 3
    squaring both sides and moving over the constant
    (s^2/343^2) + 2s/g - 9 = 0

    solving this quadratic I get s = 44.02m or -24054m

    The answer is 40.7m. Where did I go wrong?
  2. jcsd
  3. Jul 13, 2016 #2


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    Check the step where you squared both sides of the equation. Note (a + b)2 ≠ a2 + b2 in general.

    It might be good to isolate the square root on one side of the equation before squaring the equation.
  4. Jul 13, 2016 #3
    Oops. Nice spot. Thanks for the help!
  5. Jul 14, 2016 #4


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    OK. Another approach would be to let ##x = \sqrt{s}## and write your equation sqrt(2s/g) + s/343 = 3 as a quadratic equation in terms of ##x##. You will then not have to square the equation. Solve for ##x## and then find ##s##.
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