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Linear drag

  1. Nov 1, 2014 #1
    1.Small ball with mass m falls from hight h. Find how velocity, acceleration and height depend on time. Air resistance is linear and depends on velocity ##F = kv##.

    2. Air resistance ##F=kv##
    where ##k## is constant.

    3. I do not know where to go after ##ma = mg - kv ##.
     
  2. jcsd
  3. Nov 1, 2014 #2

    Mark44

    Staff: Mentor

    a = dv/dt and v = ds/dt
    Do you know anything about solving differential equations?
     
  4. Nov 1, 2014 #3
    That is the problem... I know that a= dv/dt but where do I go then? I tried to integrate but got nowhere.
     
  5. Nov 1, 2014 #4

    Mark44

    Staff: Mentor

    Show me what you did...
     
  6. Nov 1, 2014 #5
    I wrote it like ## (dv)/(dt) + v(k/m)= g##
    And then multiplied by ##e^{∫(k/m)dt}##
    to get
    ##v= e^{-(∫(k/m)dt)}(∫ge^{∫(k/m)dt}dt + c)##
    and that gave something wrong.
     
  7. Nov 1, 2014 #6

    Mark44

    Staff: Mentor

    Why don't you simplify ##e^{\int (k/m)dt}##? There's no point in leaving it like this.
     
  8. Nov 1, 2014 #7
    I simplified but did not write it here... Anyways the result i should get is ##v(t)= ((mg)/k )(1-e^{-((kt)/m))}) ## and I cannot get it. Do you see mistake ??
     
  9. Nov 1, 2014 #8

    Mark44

    Staff: Mentor

    Your work should include a line like this:
    $$ve^{(k/m)t} = g\int e^{(k/m)t}dt = g(m/k)e^{(k/m)t} + C$$
    Now solve for v.
     
  10. Nov 1, 2014 #9
    Thanks.
    But I still can't get right result? And I always get that C hanging there?
     
  11. Nov 1, 2014 #10

    Mark44

    Staff: Mentor

    What do you get when you solve for v?
     
  12. Nov 1, 2014 #11
    ##v= (g/m)k + Ce^{-((kt)/m)}##
     
  13. Nov 1, 2014 #12

    Mark44

    Staff: Mentor

    How did g(m/k) become (g/m)k? The other term looks fine.
     
  14. Nov 1, 2014 #13
    It's typo. I dont know why the solution is wrong then.
    Thanks for help!
     
  15. Nov 1, 2014 #14

    Mark44

    Staff: Mentor

    There are initial conditions that aren't being used -- at t = 0, s(0) = h and v(0) = 0. The initial velocity isn't given explicitly, but I am inferring it from the statement in the 1st post - small ball falls from height h.

    Use v(0) = 0 to get rid of the constant C. Otherwise, the solution is correct, which I've checked, and you should as well.
     
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