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Motion in one dimension

  1. Sep 27, 2009 #1
    A rock is dropped from rest into a well. The sound of the splash is heard 2.40 s after the rock is released from rest.

    a)How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s.

    (b) What If? If the travel time for the sound is neglected, what percentage error is introduced when the depth of the well is calculated?





    t=2.4 s
    V0=0 m/s
    v=336 m/s

    h= -(1/2)*9.8*(2.4-v/336)^2 m

    this the equation I have got. and when i solve it I get two values for h. but both of them have same signs. I am really lost...please help me out.
     
  2. jcsd
  3. Sep 27, 2009 #2
    How did you arrive at that equation? And what does it mean? I see two variables in it (h, v) so I can't understand it.

    Separate the motion into two regimes. One of the rock falling to the water, and the other of the sound wave traveling up from the well (You can ignore the effect gravity has on the sound wave)
     
  4. Sep 27, 2009 #3
    oh yeh i did a mistake...

    it should be h= -(1/2)*9.8*(2.4-h/336)^2 m

    yeah...I took
    h=v0t-(1/2)gt1^2...(1)

    for sound h=v0t2

    h=336*t2

    t2=h/336

    t=t1+t2

    t1=t-t2

    t1=2.4-h/336
     
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