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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?

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    $\begingroup$ 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. $\endgroup$
    – Toffomat
    Feb 17, 2022 at 10:32
  • $\begingroup$ 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? $\endgroup$ Feb 17, 2022 at 11:18
  • $\begingroup$ @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. $\endgroup$ Feb 17, 2022 at 11:32
  • $\begingroup$ Yes but in my last comment I wasn't talking about this specific problem, I wanted to know how to behave in general $\endgroup$ Feb 17, 2022 at 11:40
  • $\begingroup$ Comment about terminology: A fall with air resistance is by definition not a free fall. $\endgroup$
    – Qmechanic
    Feb 17, 2022 at 13:14

2 Answers 2

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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 $b v_y$.

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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.

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