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Find distance using speed of sound

  1. Apr 27, 2008 #1
    Two questions:
    1. A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor directly below 2.02 s later. How deep is the ocean at this point?

    I assumed this used distance = speed x time. I knew that speed must equal the speed of sound in water, which I think is 1500 m/s (pretty sure, but not 100%) and I also divided my final distance by 2 so that I only got the distance of depth one way. However, the computer (this is online homework) said I was wrong, so I'm not sure where I went wrong. Any help would probably help me with the next problem too.

    A rescue plane flies horizontally at a constant speed searching for a disabled boat. When the plane is directly above the boat, the boat's crew blows a loud horn. By the time the plane's sound detector perceives the horn's sound, the plane has traveled a distance equal to one half its altitude above the ocean. If it takes the sound 2.17 s to reach the plane, determine the altitude of the plane. Take the speed of sound to be 350 m/s.

    Just like last time, I assumed that I should use

    distance= speed of sound x time

    with a new speed (350 m/s), new time, and this time distance doesn't have to be divided by 2. However, that got me the wrong answer.

    Then I tried drawing a triangle, (since the plane has moved) and called my final distance r^2 and my x^2 = one half its altitude, so all that was left was y^2. This also was incorrect.

    Where am I making a mistake?
  2. jcsd
  3. Apr 27, 2008 #2


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    Well, what WERE your answers? How can we see where you went wrong if you don't show us what you got?
  4. Apr 27, 2008 #3
    sorry, I am new to this forum.

    for the first one I got 1515m
    343m/s x 2.02 s = 3030 m
    3030m /2 = 1515 m
  5. Apr 27, 2008 #4
    second question:
    Note: problem says : Take the speed of sound to be 350 m/s.

    attempt 1:
    distance = speed x time
    d = 350 x 2.17 s = 759.5 m

    attempt 2: using a triangle and pythagorean's theorem

    759.5 = r^2
    x^2 = "distance equal to one half its altitude above the ocean"
    x^2= (sqrt (759.5) /2)^2
    Y^2 = r^2-x^2
    y^2= 759.5-189.9
    y^2= 23.9m
  6. Apr 27, 2008 #5


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    Check your arithmetic here.
  7. Apr 27, 2008 #6
    sorry i meant speed of sound
    1500 x 2.02
  8. Apr 27, 2008 #7


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    How did you get the 189.9? If you don't know the altitude, how can you know half the altitude?? You can express the two sides of the triangle in terms of the altitude, h (what you are trying to find). Yes, 759.5 m is the length of the hypotenuse of your triangle, but make sure you are using the Pythagorean theorem correctly (be careful of your algebra).
  9. Apr 27, 2008 #8

    759.5 = (h/2) ^2 + h^2
    759.5 = (h^2 / 4) + (h)^2
    759.5 = (h)^2 (1+1/4)
    759.5 = h^2 (1.25)
    759.5 / (1.25) = h^2
    h^2 = 607.6
    h = 24.6

    Still incorrect according to the computer
  10. Apr 27, 2008 #9


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    Don't forget to square the 759.5 meters.
  11. Apr 27, 2008 #10
    awesome! I got it,

    final answer = 679.3

    Thanks for your help!

    I don't think I can use this method for the boat question, right? Do you have any suggestions for that?
  12. Apr 27, 2008 #11


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    BTW, one of the things you hopefully are learning/practicing is how to check your answers intuitively. When you found the answer, did you stop and ask yourself "does it make sense that the airplane was flying less than 25 metres above the ocean?"
  13. Apr 27, 2008 #12
    I still don't understand why

    (1500 m/s) x (2.02s) x (.5) = 1515 m

    is incorrect for the boat problem
  14. Apr 27, 2008 #13


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    Frankly, I don't understand how an online service deals with margins of error. What if the correct answer is 1514?
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