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Packaged dropped from Hot Air Balloon

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A Hot air balloon is traveling vertically up-
    ward at a constant speed of 2.6 m/s. When
    it is 19 m above the ground, a package is
    released from the balloon.
    The acceleration of gravity is 9.8 m/s^2.
    After it is released, for how long is the
    package in the air? Answer in units of s.


    2. Relevant equations
    [​IMG]

    3. The attempt at a solution
    I tried doing X = .5at^2 + vit and using -2.6, but it didnt work.
     
  2. jcsd
  3. Oct 18, 2007 #2

    learningphysics

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    Homework Helper

    you can solve the problem directly with:

    d = v1*t + (1/2)at^2

    what is d? what is v1? what is a?
     
  4. Oct 19, 2007 #3
    well a is -9.8 because of gravity.

    Vi is 2.6, but its the opposite direction of falling. So what do I do with it?

    d is 19 m.

    I dont know how to use that equation because the Velocity is upward, but the object is falling DOWN.
     
  5. Oct 19, 2007 #4

    learningphysics

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    Homework Helper

    d is -19m. becuase the object drops a distance of 19m. ie: vertical displacement = final height - initial height = 0m - 19m = -19m.

    plug in a = -9.8, vi=2.6 and d = -19 into the equation...

    solve for t.
     
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