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Wishing Well (sound wave) problem

  1. Apr 30, 2007 #1

    When you drop a rock into a well you hear the splash 2.8 seconds later.
    How deep is the well?

    2. Relevant equations
    t = t1 + t2
    d = .5 * g * (t1)^2
    d = v * t2

    3. The attempt at a solution
    since the time is so small i used the the second equation
    d = .5 * (9.81) (2.8)^2 = 38.45 m
    but when i put it into my webassign online hmwk page it said it was wrong but within 10% of the correct answer.

    does anyone have any advice. I wasnt sure how to take into account the speed of sound.
    Last edited: Apr 30, 2007
  2. jcsd
  3. Apr 30, 2007 #2


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    Homework Helper

    You need to use all of your equations. Yes, you need to take into account the speed of sound, because it relates how long it takes the sound to travel up the well. One way to approach this is to set your second and third equation equal to each other, since d is the same for both. Get the equation in terms of one of the time intervals, t1 or t2, using your first equation. You will then have to solve a quadratic. Once you get that time interval, you can find the depth of the well.
  4. Apr 30, 2007 #3

    awesome!! thanks so much for the help, i got the answer which is 35.653 m.
  5. Apr 30, 2007 #4


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    You're welcome. Good work!
  6. Jan 16, 2009 #5
    Hi! I just asked a similar question to this about another question where a rock is thrown down into a well, but this qustion was the next question in my textbook, and I still had a tiny problem too!! These well problems don't like me ;).

    I have pretty well the exact same wording in the question, except that after you drop the rock, the splash is heard 1.50 sec later. I was actually able to figure it out all on my own and get a reasonably close answer to the one in the back of the book (I got 10.6m and the answer is 11m). However, I'm confused about one thing. When I did the problem initially, I worked through the quadratic pretty quickly and just happened to leave out the -'ve sign for gravity. I got a trock drop = 1.469 sec. This was logical to me because it would take that much time for the rock to drop, leaving 1.5 - 1.469 = 0.031 sec left for the echo. So I continued and got 10.6m as the well depth which is very close to the answer which was fine.

    So just now I was checking back over it and saw that I never used gravity as -9.8 (I just left it as +9.8). So when I worked out trock drop in the quadratic with negative gravity, I got the t to be 1.533 sec. This shouldn't be possible though because that time is now over 1.5 seconds (which is the total time). When I continued to solve, I got the well depth to be 11.5m. So both of my answers are somewhat correct in terms of the answer in the back of my book, except that I'm just now sure if I'm doing something wrong with the gravity.

    This is how I am solving it:

    t1 = time for rock to drop
    t2 = time for echo

    ttotal = t1 + t2
    t2 = ttotal - t1
    t2 = 1.50 - t1

    ddrop = v0t1 + 0.5at12 (v0 is just 0m/s though, so that cancels out)
    ddrop = 0.5at12

    decho = vsound * t2
    decho = vsound * (1.50 - t1)
    decho = 343 * (1.50 - t1)
    decho = 514.5 - 343t1

    ddrop = decho
    0.5at12 = 514.5 - 343t1
    (0.5)(-9.8)t12 = 514.5 - 343t1
    -4.9t12 = 514.5 - 343t1
    0 = 4.9t12 - 343t1 + 514.5

    ---quadratic equation---

    t1 = 1.533sec and therefore, after substituing into equation (3), decho = 11.5m

    or, if I used g = +9.8 (instead of -9.8) in my quadratic equation, I get:
    t1 = 1.469sec and therefore decho = 10.6m

    I hope that makes some kind of sense? I think I may just have a sign convention switched around or something. I've checked over it a few times, but can't seem to find a problem. I don't think the answers are too far off, but I'm just curious as to why I got the answer with a +g instead of a -g. Seems backwards to me!! Thanks!
  7. Jan 16, 2009 #6


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    The sign of gravity here is not relevant.

    The distance is determined by the magnitude or 9.8, not its direction.

    Your equation should be 4.9t² + 343t - 514.5 = 0

    This should yield 1.469 sec for the trip down - your second answer.

    The first one of course isn't possible since it is longer than it took to hear the splash.
  8. Jan 16, 2009 #7
    Oh! So the direction of gravity is not even used! Alright. I guess that makes some sense to me in that we're only finding a magnitude of distance as well and we're not concerned with direction. I'm just kind of leary on leaving out that negative because I've taught myself that using sign convention every time for all numbers will make the problem work. And so far it has, but I guess I'll have to make an exception, lol!! Thanks so much for your help LowlyPion!

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