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The echo

  1. May 9, 2007 #1
    A stone is dropped into a deep canyon and is heard to strike the bottom 13.4 s after release (the speed sound is 343m/s) What would be the percentage error in the depth if the time required for the sound to reach the canyon rim were ignored?

    I found the depth of the canyon =2298m
    then what is the next step?
     
  2. jcsd
  3. May 9, 2007 #2

    Doc Al

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    How did you do calculate the depth?

    The point is to calculate it two ways: once ignoring the time for the sound to travel; once taking it into account.
     
  4. May 10, 2007 #3
    sorrry
    the distance = 1/2 a t^2
    = .5 x 9.8 x (13.4/2)^2 = 219.961 m^2


    that is when i take the time into account

    but how i will not take it??
     
    Last edited: May 10, 2007
  5. May 10, 2007 #4

    Doc Al

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    The time needed here is the time that the stone was falling. Why did you divide by 2?

    Think of the time to hear the echo as having two parts:
    (1) The time that the stone was falling.
    (2) The time it takes for the sound of the stone hitting the canyon floor to travel up to the top.

    Time #1 + Time #2 = 13.4 seconds

    Since sound travels pretty fast compared to the stone, time #2 will be small.
     
  6. May 10, 2007 #5
    ok that is very good

    then how i will divide the time between the falling and the echo
     
  7. May 10, 2007 #6

    Doc Al

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    To incorporate the time delay due to the speed of sound, you'll need to set up equations and solve for the times. Hint: Set up two equations (one for the falling rock; one for the sound) relating distance and time.
     
  8. May 11, 2007 #7
    ok

    d= .5xgxt1 = 343t2
    t1+t2 =13.4
    by solving both equations
    t1= 11.51
    t2= 1.89

    then the depth = 648.9 m

    is it right


    next step:

    %=(depth1-depth2)x100/(depth1)

    we found one depth

    the second depth when we ignored the time
    is the second depth when time = 13.4 for the d=.5x9.8xt
     
  9. May 11, 2007 #8

    Doc Al

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    Looks good to me.
    Right. (You have a typo in your formula--that should be t^2.)
     
  10. May 11, 2007 #9
    thanks

    i posted it ...but it is wrong!!!!

    (879.844-648.9)/(879.844)=26.2% xxx Wrong xxx !!

    why??
     
  11. May 11, 2007 #10

    Doc Al

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    You need to read the question very carefully. I think they are asking for the percentage error the incorrect method (that ignores the sound travel time) introduces with respect to the correct method. The correct measurement should be in the denominator.
     
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