# Is air drag equation in term of momentum still valid?

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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$$\int \frac {dv_y}{(1-\frac {bv_y}{mg})}=g\int dt$$

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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I don't think you really answer his question here. –  Bernhard Dec 15 '12 at 13:39