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One-dimensional motion with drag

  1. Nov 11, 2013 #1
    Hi, this is my first time posting on this forum. This is a question about how to approach questions about one-dimensional motion with drag.

    A link to the question:
    http://puu.sh/5fOCO.png [Broken]

    While attempting this question, I thought the two equations they were referring to were:
    dy/dt = v
    dv/dt = -g

    However, this would be far too basic for 4 marks (this is a past exam).

    It would be much appreciated if you could help me out with this question.

    Thanks.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 11, 2013 #2
    After some thinking, I got:

    ma = -sign(v)*F_d - mg

    So, a = -sign(v)*F_d/m - g

    And of course I can sub in F_d which is given in the question. Not sure about how to get dy/dt though.
     
  4. Nov 12, 2013 #3

    ehild

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    Homework Helper
    Gold Member

    Fd is proportional to v2, say Fd/m=b.

    You need to determine the functions v(t) and y(t), by solving the differential equation

    dv/dt = -sign(v)*b v2 - g,

    and then integrating v(t) = dy/dt to get y(t).


    ehild
     
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