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Free fall w/ Air Resistance

  1. Feb 15, 2012 #1
    A ball of mass m is released from rest at x=0. Air resistance is expressed as R = kv2 , where k is a positive constant and v denotes velocity.

    Derive an expression for the speed in terms of the distance x that it has fallen.
    Identify the terminal v.

    I know I have to start with ma = mg - kv2 and integrate but I can't seem to separate the variables right to get it as function of x or distance (even though the problem has a picture of a vertical fall)
     
  2. jcsd
  3. Feb 15, 2012 #2
    Hint: Express the 'a' in your equation differently.
     
  4. Feb 15, 2012 #3
    Okay so I have mv(dv/dx) = mg - kv2 , and divide both sides by the (mg - kv2)

    Then, (mv/(mg - kv2))dv = dx

    ∫(mv/mg - mv/kv2) dv = ∫dx

    Just need a little help integrating if I'm on the right track I guess....
     
  5. Feb 15, 2012 #4
    Acceleration is not the derivative of velocity with respect to distance; it is the derivative with respect to time.
     
  6. Feb 15, 2012 #5
    Great....just realized that
     
  7. Feb 15, 2012 #6
    If I change a to d2s/dt2 is this the right path?
     
  8. Feb 15, 2012 #7

    HallsofIvy

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    Since your force is given in terms of v, it would be better to write the acceleration as dv/dt:
    [tex]m\frac{dv}{dt}= mg- kv^2[/tex]
    and there is no problem with "separating variables"- you have
    [tex]\frac{mdv}{mg- kv^2}= dt[/tex]
     
  9. Feb 15, 2012 #8
    But I need the velocity in terms of the distance fallen, x. Integrating the right side, ∫dt , would just give me 't' am I right?
     
  10. Feb 15, 2012 #9
    Was I not correct in saying that a = v (dv/ds) ??
     
  11. Feb 15, 2012 #10
    In my book it says v dv = a ds --> a = v(dv/ds).....

    so I have vdv/(-g + kv2) = dx

    ∫vdv/(-g + kv2) [from 0 to v] = ∫dx [from 0 to x]

    (1/2k) ln(-g+kv2) [from 0 to v] = x

    --> (1/2k) ln(g-kv2/g) = x

    Solving for v and skipping a few steps I got v = √(g/k(1-e2kx))
     
  12. Feb 15, 2012 #11
    Your units do not show velocity. You need a mass in there.
     
  13. Feb 15, 2012 #12
    You're right....should start as dx = mvdv/(-mg + kv2)
     
  14. Feb 15, 2012 #13
    Whenever you derive something, it is always prudent to check your units in the final product to see that they are what is intended. Secondly, it is a good idea to explore the limits of your formula to see if it makes sense at its limits.
     
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