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Homework Help: Free fall with air resistance, must find velocity

  1. Jan 3, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is to find the velocity as a function of time, given the following;
    the object is in free fall and its initial velocity is zero. The formula given for the air resistance is: a = g - kv, where g is acceleration d/t gravity, k is a constant, and v is velocity.

    2. Relevant equations
    The problem gives the hint to solve using integration by parts, with u = g - kv

    3. The attempt at a solution
    The solution given in the back of the book is v = [itex]\frac{g}{k}[/itex](1-e-kt)
    I have no idea how to arrive at this solution.
  2. jcsd
  3. Jan 3, 2012 #2


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    [itex]g - kv[/itex] is the expression for the net downward acceleration of the object, not just the deceleration due to wind resistance.

    Set up the differential equation relating velocity and time, to begin with.
  4. Jan 3, 2012 #3
    You can use differential form of acceleration, ie [itex]\large{a = \frac{dv}{dt}}[/itex]
  5. Jan 4, 2012 #4


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    You don't have to integrate by parts, BTW. This is an easily separable first order differential equation. Maybe a simple substitution to clarify the form, but that's it.
  6. Jan 4, 2012 #5
    Thanks for the replies everyone, but I'm still not seeing it. I am not trying to copy and paste an answer here, I'm doing this on my own time, trying to study before I go back to school in the Fall.
    Where does the e come from? Is there some basic calculus that I'm overlooking? It looks to me that solving the problem should go something like this:

    [itex]\frac{dv}{dt}[/itex]= g - kv

    ∫[itex]^{t}_{0}[/itex]dvdt = ∫[itex]^{t}_{0}[/itex](g - kvt)dt

    v = gt - kvt → v = [itex]\frac{gt}{1 + kt}[/itex]

    Which makes sense, but doesn't match the answer in the book. Is the book incorrect, or am I? What am I missing here?

    Thanks again for the help everyone.
  7. Jan 4, 2012 #6
    When you neglect the gravity term, you should immediately see that the solution of
    because of the property of exponential functions:

    Look up how to solve first order ode's using integrating factors and solving homogeneous and non-homogeneous equations. Hope this helps.
  8. Jan 4, 2012 #7
    Why would you neglect the gravity term?
  9. Jan 4, 2012 #8


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    This integration is wrong. v is a function of t and you cannot treat it as a constant as you did on the right.

    [tex]\frac{dv}{dt}= g- kv[/tex]
    you get
    [tex]\int\frac{dv}{g- kv}= \int dt[/tex]
    to integrate on the left, let u= g- kv so that du= -k dv, dv= (-1/k)du
    [tex]-\frac{1}{k}\int\frac{du}{u}= \int dt[/tex]
    so that
    [tex]-\frac{1}{k}ln u= ln(u^{-1/k})= t+ c[/tex]
    where c is the constant of integration. Taking the exponential of both sides
    [tex]u^{-1/k}= e^{t+ c}= Ce^t[/tex]
    Take the -k th power of both sides to get
    [tex]u= g- kv= Ae^{-kt}[/tex]
    where [itex]A= C^{-k}[/itex]

    Finally, solve for v:
    [tex]v(t)= \frac{Ae^{kt}- g}{k}[/tex]
  10. Jan 4, 2012 #9
    Thank you very much Halls, that was a huge help. (It is a definite integral, though, evaluated at an arbitrary time t and assuming v=0 at t=0. I did away with the constant of integration.) I still can't get the answer that is given in the back of the book, however. When I manipulate it to a form similar to the answer above, I get v=[itex]\frac{g}{k}[/itex](1-[itex]\frac{1}{g}[/itex]e-kt) (the exp should not be divided by g). Another strange thing is that both the answer I got using your method and the answer in the book check out when you differentiate with respect to t and set the result equal to a=g-kv, so both appear to be valid, but that can't be right. Any ideas?
  11. Feb 26, 2012 #10

    How would you solve differently if you had m(dv/dt)= mg- kv2
  12. Feb 27, 2012 #11
    [itex]\int \frac{1}{1-ax^2}dx=\frac{1}{\sqrt{a}}arctanh(\sqrt(a)x)[/itex]

    There is probably a nice variable transformation to get an easier intermediate integral in case you don't know the above integral.
  13. Feb 27, 2012 #12
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