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Simple problem. Depth of well providing sound reaches you in 1.5 second.

  1. Apr 8, 2004 #1
    I was wondering you someone could help me with an alternative way of solving this problem. Just to help my algebra a bit.
    You drop a rock down a well. Find the depth of the well providing the sound of the splash reaches you in 1.5 seconds.

    First off, we know speed of sound is 343 m/s.
    Here are my equations:
    t1+t2=1.5 s t1 is time for rock to hit bottom t2 is time for sound to reach your ears.
    x=.5*a*t^2 -> x=4.9*t1^2
    v=x/t -> x=343*t2

    Now I sub the last 2 equations to get:
    343*t2=4.9*t1^2
    Now we sub. for t2:
    343*(1.5-t1)=4.9*t1^2
    From here we can reduce, and solve with the quadratic formula.

    I was wondering how you would solve the equation, if you sub. for t1 instead of t2. This would make the equation:
    343*t2=4.9*(t2-1.5)^2
     
  2. jcsd
  3. Apr 8, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    It's just another quadratic equation. Multiply out the term on the right (do the square) and put it into standard form.
     
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