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Freefall problem

  1. Sep 20, 2004 #1
    Ok I am at this problem for hours and I am still nowhere, I am just wondering if any of you can help.

    Here is the question

    A rock is dropped from rest into a well. If the sound of the splash is heard 2.42s later, how far below the top of the well is the surface of the water? (take the speed of sound to be 336.0 m/s and take the freefall acceleration of the rock to be 9.8 m/s^2.

    So I figure ok this must mean

    2.42s = Time spent falling + Amount of time it took for the sound to reach the top.

    So I try to use one of the kinematic equations for motions with constant acceleration

    Sf = Si + Vi*(Change of T) + 1/2 * A *(Change of T) ^ 2

    Since Initial Position Si = 0 and Vi = 0 The formula will look like

    Sf = 1/2 *A*(Change of T) ^ 2

    Since I need to get rid of one of the unknowns I do this

    Sf = 1/2 * 9.8 * (2.42s - SF/336m/s) ^ 2

    Sf = 4.9 * (2.42s- SF/336m/s) ^ 2

    Sf = 4.9 * (813.12m - SF)/336m/s)^ 2

    And then yeah..I went further then that but let's just say it didn't give me the right answer.

    I tried several other methods but nothing gave me the right answer.

    Any Ideas on how to resolve this matter?
  2. jcsd
  3. Sep 20, 2004 #2
    It seems that you use only one change of time but there are two times t1 fall time, t2 is time for the sound to get back up. And t1+t2=2.42

    For the distance we know that S=0,5g*t1 but also S=Vsound*t2 and

    So now you can make youre square equation with standard solutions.
  4. Sep 21, 2004 #3
    Sorry, I have been out of school for one year, and I now have no idea how to make square equation with standard solutions.

    Can you provide an example please?
  5. Sep 22, 2004 #4
    come on guys, I really need some help here.
  6. Sep 22, 2004 #5
    Here's how I would do it.
    Let T1 = time for rock to fall
    Let T2 = time for sound to reach you

    We know from freefall :
    X = (1/2) a T^2

    X = (1/2) a (T1)^2

    Then from the sound reaching you we have:
    X = Velocity * Time

    X = V*T2

    combine that with first as X=X

    V*T2 = (1/2) a * (T1)^2

    We also know T1+T2 = 2.42 seconds
    T2 = 2.42 - T1
    336*(2.42-T1) = (1/2)*9.8*(T1)^2
    Write as a quadratic equation :
    (multiply things out, then get T1^2 alone, etc.

    T1^2 + 68.6 T1 - 165.9 = 0

    Use solution of quadratic equations to get Ti, then plub into x=(1/2)(9.8)T1^2 to get X. (search google if you dont know solution of quadratic).

    thats how I would do it.
    I get 26.5 meters.
  7. Sep 22, 2004 #6
    Wow, you gave a really easy to follow explanation and that answer matches with the answer from the back of the book as well.

    Thank you very much.
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