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DE: Sound more like an algebra problem :x

  1. Oct 4, 2008 #1
    A stone is released from rest and dropped into a deep well. Eight seconds later, the sound of the stone splashing into the water at the bottom of the well returns to the ear of the person who released the stone. How long does it take the stone to drop to the bottom of the well? How deep is the well?

    I pretty much looked at the solution manual.

    [tex]d=4.9t^2[/tex] How does that come into play?

    [tex]d=340s[/tex] distance = the rate of the speed of sound x the splash?

    [tex]s=8-t[/tex] So when t=8, splash is 0?


  2. jcsd
  3. Oct 4, 2008 #2
    [tex] h= \frac{1}{2}g t_a^2[/tex]

    [tex] v_{sound} \cdot t_b = h [/tex]

    [tex] t_a + t_b =8 [/tex]

    Do you get this?
  4. Oct 4, 2008 #3
    What does a and b represent? t_a ... t_b
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