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A rock dropped from a cliff using speed of sound

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 5.1 s later. If the speed of sound is 340 m/s, how high is the cliff?



    2. Relevant equations
    d=v/t(didnt work)



    3. The attempt at a solution
    My first attempt at this problem was to try that formula above, but it was not the correct way to do this problem. I then realized that the time i had was the total time, and that i needed to find the time it took for the rock to hit the water, or the time it took for the speed of sound to go up, which i need distance for. I am now stuck there :(
     
  2. jcsd
  3. Aug 31, 2009 #2

    Hootenanny

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    Welcome to Physics Forums.

    You are indeed correct that we first need to determine the time taken for the rock to fall. How do you suppose that we could work that out?
     
  4. Aug 31, 2009 #3

    Maroc

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    From all i know is that you have to divide your distance by 2. Because you are calculating it from bottom to top.
     
  5. Aug 31, 2009 #4

    Hootenanny

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    And why would we want to divide it by two?
     
  6. Aug 31, 2009 #5

    Chronos

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    Not so simple, rock drops due to gravitational acceleration whereas sound travels at the speed of sound. Division by two is not the answer. Integrate and compare the time lag. 5.1 seconds sounds suspiciously close.
     
  7. Aug 31, 2009 #6
    Hi there,

    You have to divided the problem in two parts. One where the rock is in free fall. Then, once the rock hits the ocean, the sound is emitted and travels at constant speed.

    Cheers
     
  8. Aug 31, 2009 #7
    This problem could be interpretted in two ways:

    1) the rock is dropped and the total time elapsed is 5.1 seconds until the sound is heard. A two part problem where the two parts will have to be equated to each other in terms of distance.

    2) the rock is dropped and the dropper sees it hitting the ocean and hears the sound 5.1s later. In this case it is simply the distance that a sound wave would travel in 5.1 seconds at 340m/s
     
  9. Aug 31, 2009 #8

    Hootenanny

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    I would say that it's quite clear that the only correct interpretation is #1. The question clearly states that the rock is dropped and then 5.1 seconds later a sound is heard. It does not say that the rock is dropped it hits the ocean and 5.1 seconds later a sound is heard.
     
  10. Aug 31, 2009 #9

    Maroc

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    oh nvm got confused with another problem. Sorry.
     
  11. Aug 31, 2009 #10
    Okay, so i took the time to look over this problem and seperated the problem into two parts. I let d be my distance, so d = (0)(t1) + 1/2(9.8)t1^2. t1 being my first time. I then took d = vt and used the speed of sound for this problem, so d = (340m/s)(t2). t2 being my second time. i then made the problems equal to each other so (340)t2 = 1/2(9.8)(t1^2). I seem to be stuck here, can anyone help me with the algebra in this problem?
     
  12. Aug 31, 2009 #11
    Getting warmer...t1+t2=5.1s, so t2=5.1-t1. I therefore can substitute in 340(t2) as 340(5.1-t1)=4.9(t1^2)
     
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