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Penny Dropped down Well Problem

  1. Oct 22, 2006 #1
    A penny is dropped down a well. One hears the sound of it hitting the bottom 2.5 seconds later. If sounds travels at 330 m/s, how deep is the well.

    What I did:
    Time(penny)+Time(sound)=2.5
    T(p)=2.5-T(s)
    T(s)=x
    T(p)=2.5-x
    d=vit+1/2at^2
    330x=1/2(-9.8)(2.5-x)^2
    I then solved using the quadratic forumula and got -62.2 or-0.102. I then chose the 0.102 seconds and multipled it by 330 and got 33.7 m. If this is right, my question is how can both times be negative, and how can you decifer which one is right besides using common sense?
     
  2. jcsd
  3. Oct 23, 2006 #2

    OlderDan

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    You have sign problems? If the penny fell for the full 2.5 seconds, how far could it fall?
     
  4. Oct 23, 2006 #3
    no the time that it fell plus the time is took for you to hear it totals 2.5 seconds.
     
  5. Oct 23, 2006 #4

    OlderDan

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    I know that. The point is that in 2.5 seconds the penny could not even fall as far as you said it did. You made a mistake in your calculations.
     
  6. Oct 23, 2006 #5
    weel can you show me what i did wrong?
     
  7. Oct 23, 2006 #6

    OlderDan

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    I don't see all your work, but it is probably here that you went wrong
    330x is the distance the sound moves, a positive number

    1/2(-9.8)(2.5-x)^2 is supposed to be the distance the penny moves, another positive number, but you made it negative. My guess is that if you get rid of the - in fornt of the 9.8 it will solve your problem.
     
  8. Oct 23, 2006 #7
    isnt acceleration always negative tho in regards to gravity? and I thought since the penny is moving down, isnt that a negative direction?
     
    Last edited: Oct 23, 2006
  9. Oct 23, 2006 #8
    I also did what you said and got 28.5 m which seems right. But for the time it takes the sound to reach the top, i got 0.086 sec or 72.26 sec. Why do you get two answers in this case? What is the significance of the 72.26 seconds?
     
  10. Oct 23, 2006 #9

    OlderDan

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    This is common when you solve quadratic equations. You often get a solution that is mathematically correct, but has no physical significance.
     
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