This is the known equation of air drag:
$$m{\bf a}=mg-\mathcal D=mg-b{\bf v}.$$
Considering this, is air drag equation in term of momentum still valid?
$$m{\bf v}=mv_g-b{\bf r}.$$
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2 Answers
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$$\int \frac {dv_y}{(1-\frac {bv_y}{mg})}=g\int dt$$
so momentum will be a function of time: $$p(t)=m.v_y(t)=\frac {m^2.g}{b}(1-{\bf e}^{\frac {-bt}{m}})$$
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2$\begingroup$ I don't think you really answer his question here. $\endgroup$– BernhardCommented Dec 15, 2012 at 13:39
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No, you cannot just change force to velocity that way.
I know, it looks like you can just "integrate" every term on its own, but if you solve the differential equation for a specific case, you'll see that things are more complicated than that.
A key thing that should signal to you that you are probably wrong is that your velocity depends on position with the b r term. While in some special cases (e.g., simple harmonic motion, gravitational orbits) you can express velocity as a function of position, it's not really the position that determines your velocity in the way that velocity determines a damping force in a fluid, or the way that your position determines the force for an electric or gravitational or spring force. So that was your first clue.
Another clue that you were doing things wrong was your vg. Did you mean vy? Either way, you can tell from the "no friction" case (ie, set b = 0) that you can't just change a gravitational acceleration to a velocity.
