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Stone-Drop Time

  1. Sep 27, 2009 #1

    jacksonpeeble

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    Gold Member

    1. The problem statement, all variables and given/known data
    A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.24 s after the stone is dropped. How deep is the hole?


    2. Relevant equations
    https://www.physicsforums.com/showthread.php?t=93392


    3. The attempt at a solution
    I attempted to follow the steps (substituting different values) in the post above; however, I do not understand the process adequately (nor did I come up with the correct answer),
     
  2. jcsd
  3. Sep 27, 2009 #2
    There are two components to this. The stone falling to the bottom of the well and the sound traveling to the top of the well. You need to make an equation for both parts and also an equation for the total time.
     
  4. Sep 27, 2009 #3
    Okay so you know [tex]d=4.9t^{2}_{1}[/tex] on the way down. On the way up [tex]d=343t_{2}[/tex] and these two distances are equal. Lastly you are given that [tex]t_{1}+t_{2}=1.24s[/tex]

    If you set the equations of motions equal to each other you are left with an equation of time. [tex]4.9t^{2}_{1}=343t_{2}[/tex]

    What could you substitute in for [tex]t_{2}[/tex] that would let you solve for [tex]t_{1}[/tex] ? Keep this relationship in mind [tex]t_{1}+t_{2}=1.24s[/tex] !
     
  5. Sep 28, 2009 #4

    jacksonpeeble

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    Gold Member

    Thank you both very much! From this, t=1.21878 (disregarding the negative value yielded from the quadratic equation.

    I then plugged in the value to the equation:
    0*1.219+.5*9.8*1.219^2=7.279
     
    Last edited: Sep 28, 2009
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