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Acceleration and distance problem

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130m above the water.

    2. Relevant equations

    Vf=Vi+ax*delta t

    (Vf)squared=(Vi)squared + 2(ax)*delta x

    3. The attempt at a solution

    I think I'm supposed to use the first equation to figure out Vi. I calculated Vf using acceleration of 9.8 *6s, which equals 58.8 m/s. But if i throw in Vf, a, and delta t, into the first equation, Vi equals 1. I dont know if this is right so far, but I'm stuck here so any help would be appreciated. Thanks.
  2. jcsd
  3. Sep 9, 2007 #2
    where do you get 6s?
    all you seem to be given is:
    y=130m above the water.

    Is there more to that problem? since you don't say what the question is exactly.
  4. Sep 9, 2007 #3
    Crap, my bad. This is the question:

    If the briefcase hits the water 6.0s later, what was the speed at which the helicopter was ascending?
  5. Sep 9, 2007 #4
    O ok, well is the 9.8 the acceleration of the brief case? because what other force was acting on the briefcase? if the 9.8 was gravity going down, was something pulling it "up"?
  6. Sep 9, 2007 #5
    i'm not really sure. if there is anything pulling it up, it would be the helicopter. that is all the information they provided us with.
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