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Motion with constant acceleration

  1. Jun 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A person is standing over a well and drops a rock. After 2.4 seconds he hears a rock splashing into the water. Take the speed of sound to be 300m/s and calculate the depth of the well.

    2. Relevant equations
    x=v/t for constant velocity
    x=1/2at^2 for accelerated motion

    3. The attempt at a solution
    I tried this solving by taking t1=x/v x is depth of well and v speed of sound and also t2=(2x/g)^1/2
    t1+t2=2.4 but i keep getting x to be 104.56 meters which cant be right although it seems all my steps are conclusive. Substituted t1 and t2 took p=x^1/2 and solved quadratic equation for p and god 10.4 m which goes to 104 m = x which is not right. What do you think is wrong?
     
  2. jcsd
  3. Jun 28, 2015 #2

    CWatters

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    As a check work the numbers backwards..

    On the way down..

    d = ut+ 0.5gt2
    u=0 so..
    t = SQRT(2d/g) = SQRT(2*104/9.81) = 4.6 seconds

    So you made a mistake somewhere.

    Show your working?
     
  4. Jun 28, 2015 #3
    T1+T2=2.4
    X/v + ((2x)^1/2)/g^1:2 = 2.4
    (g^1/2)x + v(2^1/2)x^1/2 = c
    Call c = 2.4vg^1/2
    take p=x^1/2
    (g^1/2)p^2 + v(2^1/2)p - c = 0
    I get p = 10.4 m
     
  5. Jun 28, 2015 #4

    PeroK

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    I think your equation [*] is correct, but you made a mistake solving it.
     
  6. Jun 28, 2015 #5
    I solved it twice getting the same answer
     
  7. Jun 28, 2015 #6

    PeroK

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    Maybe you made the same mistake twice. You didn't forget something in the denominator of the formula for quadratic roots by any chance?
     
  8. Jun 28, 2015 #7
    Oops 2a, not a...
     
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