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Help with kinematics in one demension free fall

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data
    A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it taktes for the ball to fall halfway to the ground is 1.2s. Find the time it takes for the ball to fall from rest all the way to the ground

    2. Relevant equations
    v=vo + at


    3. The attempt at a solution
    0 = 4.6 + (-9.80)t
    t = - 5.2 s. time cant be negative so 5.2s
     
  2. jcsd
  3. Sep 11, 2008 #2

    rl.bhat

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    Use the equation h = vo*t + 1/2*g*t^2
    Using vo = 0 and t = 1.2s find the half height and then full height. From that find the time to fall from rest to the ground.
     
  4. Sep 11, 2008 #3
    i dont understand why cant i use v=vo + at
    and did i mess up on the time?
     
    Last edited: Sep 11, 2008
  5. Sep 11, 2008 #4
    0 times 1.2 + .5 (-9.80)(1.44) = -7.056 gives me the height. full height is 7.056 times 2 = 14.112
    now i am stuck again >.<

    wait... now i use v=vo + at to find the final so

    0 + (-9.80)1.2 = 11.76 = final velocity

    then i use y = 1/2 ( v + v )t

    y = 1/2(0 + -11.76)1.2
    y = -7.056 omg i got the same thing?!?!?
     
    Last edited: Sep 11, 2008
  6. Sep 11, 2008 #5

    LowlyPion

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    Maybe look at it this way:

    The distance dropped from rest is equal to 1/2 a*t2.

    Exploiting this observe then that

    [tex]\frac{H_1}{H_2} =\frac{1/2*g*t_1^2}{1/2*g*t_2^2}[/tex]

    This simplifies to - plugging in that H1 = H , and H2 = 2H

    [tex]\frac{H}{2H} = \frac{1.2^2}{t_2^2} = \frac{1}{2}[/tex]

    [tex]t_2 = 1.2 * \sqrt{2}[/tex]

    Evaluating this equation looks remarkably easier to me.
     
  7. Sep 11, 2008 #6
    ahh thank you
     
  8. Sep 11, 2008 #7

    LowlyPion

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    So long as you learn the technique, you're welcome.

    Good luck.
     
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