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Motion of a particle subject to a variable resisitive force

  1. Feb 28, 2009 #1
    http://img338.imageshack.us/img338/193/33379676.th.jpg [Broken]

    See the image above. I am able to get to (where x = distance):

    x = (1/2k)ln((g - k(V^2)(tan a )^2)/(g - kv^2))

    If you let v = 0, and then multiply by mg, you find potential energy gained presumably - I dont understand why the question states the potential energy lost.

    Obviously the 0.5(MV^2)*(tan a)^2 part of the expression is due to the lost of the kinetic energy, however for the other part of the expression I am struggling.

    Thanks for any help in advance.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 28, 2009 #2
    If the stone is thrown vertically upwards how can the terminal velocity exist?
     
  4. Mar 1, 2009 #3
    ah right, so why does it say in the ascent, as opposed to descent?

    Thanks
     
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