1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    therefore
    +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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Kinematics/speed of sound question
Loading...