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Homework Help: Velocity of a wrecking ball

  1. Nov 1, 2006 #1
    A wrecking ball suspended from a 12.0m cable is accelerated up to 5.00m/s at it's lowest point. After that point the ball has negative aceleration as it gains elevation. What is the ball's velocity when the cable makes a 20.0 degree angle with the verticle?

    Is this just a matter of finding the height of the ball at that point and calculating the velocity at that height? I guess I could use the potential energy formula (U=mgh) some how. I'm just not sure on this one.
     
  2. jcsd
  3. Nov 1, 2006 #2
    I like the U=mgh idea; you know what the energy of the system is because your given the speed (therefore the kinetic energy) at the lowest point.
     
  4. Nov 1, 2006 #3

    AVD

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    i dont understand are you saying that its velocity at the lowest point is 5.00m/s ? If so try using conservation of energy.
     
  5. Nov 1, 2006 #4
    Oops, in my initial post, I meant to say that the ball was accelerated to 5.00m/s, so yes, the velocity at the lowest point is 5.00m/s.
     
  6. Nov 1, 2006 #5
    OKay, then your on the right track. Use conservation of energy. Set up the equation like this:
    KE=KE+mgh so...
    1/2mVi^2=1/2mVf^2+mgh, where Vi=initial velocity=5m/s, and Vf=unknown
    The mass cancels out, your left with:
    1/2Vi^2=1/2Vf^2+gh
    use trig to find h at 20degrees, then plug in your values to find final velocity.
     
  7. Nov 1, 2006 #6
    Thanks, got it!

    V=sqrt(((5.00m/s)^2)-2(9.8)(12.0-12.0cos20.0))=3.29m/s
     
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