# Differential equation of the free falling object with air resistance

I'm puzzled by the result I get when I try to solve the differential equation for an object that is falling subject to air resistance. Suppose the $$y$$-axis is directed upwards, then $$\vec{F_a}=bv \hat{y}$$ and $$\vec{P}=-mg\hat{y}$$, where $$b,v, g$$ are positive. This leads me to the following equation: $$\\m\frac{dv}{dt}=bv-mg,$$ which eventually leads to $$v(t)=\frac{mg}{b}(1-e^{bt/m})$$ which is obviously an absurd result since we expect $$\lim_{t\rightarrow \infty}v(t)=mg/b.$$ What am I missing? Are the forces involved described correctly in my frame of reference?

• You have missed a minus sign: Your velocity is the downward velocity ($\vec v =-v \hat y$), so there is an extra minus sign on the lhs of your differnetial equation. Then, you get $v(t)=\frac{mg}{b}(1-e^{-bt/m})$ as expected. Commented Feb 17, 2022 at 10:32
• Uh, now I understand, thanks. One more thing: in cases where I don't know the orientation of the velocity, what should I do? Should I try both $m\frac{dv}{dt}$ and $-m\frac{dv}{dt}$ and see which one leads to a physically acceptable result? Commented Feb 17, 2022 at 11:18
• @GabrielePrivitera the drag force is always opposite in direction to the relative velocity of the object with respect to the air. You could use sgn(v) as appropriate. Commented Feb 17, 2022 at 11:32
• Yes but in my last comment I wasn't talking about this specific problem, I wanted to know how to behave in general Commented Feb 17, 2022 at 11:40
• Comment about terminology: A fall with air resistance is by definition not a free fall. Commented Feb 17, 2022 at 13:14

You have missed a minus sign: Your velocity is the downward velocity ($$\vec v =−v\hat y$$), so there is an extra minus sign on the lhs of your differnetial equation. Then, you get $$v(t)=mgb(1−e−bt/m)$$ as expected.

In general, you can write $$\vec v= v_x \hat x +v_y \hat y$$) and solve the equation of motion ($$m\dot{\vec v}=\vec F$$) with the appropriate boundary conditions. In your case, you assume $$v_x(t=0)=0$$, and so $$v_x$$ stays zero; for $$v_y$$ the main point is that you have to keep track of the relative signs between the $$\dot v_y$$ term and the drag force term $$b v_y$$.

Suppose the y-axis is directed upwards, then $$F_a→=bv\hat{y}$$ and $$P=−mg\hat{y}$$, where b,v,g are positive. This leads me to the following equation:

The bold part is where the entire problem lies. You decided to make the positive direction of y upwards, right? But you are describing a falling motion, this means that velocity will be negative, right? But you said that b is positive, but if b is positive, then $$F_a→=bv\hat{y}$$ will be negative and because $$v$$ is negative and is being multiplied by a positive number $$b$$. So, the drag force will be negative, pointing in the same direction of the weight, which is incorrect.

Your equations: $$m\frac{dv}{dt} = bv - mg$$

$$v(t) = \frac{mg}{b}(1-e^{\frac{bt}{m}})$$

Are absolutely correct, but you are using b as a positive number, so the limit doesn't converge to the terminal velocity. But assume that $$b$$ is negative and voila, the $$e^{\frac{bt}{m}}$$ converges to 0 as $$t$$ goes to infinity. And the terminal velocity goes to $$v_{ter} = \frac{mg}{b}$$. And, since $$b$$ is negative, again, this velocity is negative, pointing downwards as defined by your axis orientation.

So, you were correct in your equation solving, your assessment of the $$b$$ constant was incorrect, in your system of coordinates $$b$$ is negative.