I'm learning about linear drag and I have a question with regards to the y component. The y component of the equation of motion (if we measure y downwards) is given by:

$$m \dot{v}_y = mg - bv_y$$

where b is just a coefficient of linear drag. From this, one can show that:

$$m \dot{v}_y = -b(v_y - v_{ter})$$

where $v_{ter}$ is the terminal velocity given by $v_{ter} = \frac{mg}{b}$ for linear drag. From this, we find $y(t)$ of the form:

$$y(t) = v_{ter}t + (v_{yo} - v_{ter})\tau(1-e^{-\frac{t}{\tau}})$$

where $v_{yo}$ is the intitial speed in the y direction of the projectile, and $\tau$ = $\frac{m}{b}$. My issue is that the book I'm using tells me that if we measure y vertically upwards, the sign of the terminal velocity in the final equation flips and becomes:

$$y(t) = -v_{ter}t + (v_{yo} + v_{ter})\tau(1-e^{-\frac{t}{\tau}})$$

I'm not sure why this is the case, because I thought that if the sign flips for the opposite direction of the y axis, then the sign for the initial speed would also change to a negative (I thought it was a case of flipping the signs of the equation of motion at the top) but it doesn't appear to do so. Could anyone point out where I'm going wrong?

  • $\begingroup$ Please provide information about the textbook which you are using. $\endgroup$
    – Farcher
    Apr 8, 2020 at 22:57
  • $\begingroup$ John Taylor, "Classical Mechanics" $\endgroup$
    – Andrew
    Apr 8, 2020 at 23:16

1 Answer 1


It is really all to do with which quantities you choose to be components (can be positive or negative) and which you choose to be magnitudes (can only be positive).

The y component of the equation of motion (if we measure y downwards - unit vector $\hat y$) is given by:

$m \dot{v}_y \hat y= m\,\vec g - bv_y \hat y$

Now what do you do with $\vec g$?

You say $\vec g = g \hat y$ and $g$ is the magnitude of $\vec g$ whereas $\dot{v}_y$ and $v_y$ are components in the $\hat y$ direction.

So the equation of motion becomes $m \dot{v}_y \hat y= m\,g\hat y - bv_y \hat y\Rightarrow m \dot{v}_y = m\,g - bv_y$ which is the equation that you quoted.

Now suppose you choose up a positive - unit vector $\hat Y$.
The equation of motion looks very similar $m \dot{v}_Y \hat Y= m\,\vec g - bv_Y \hat Y$ but now you would write that $\vec g = - g \,\hat Y$ where again $g$ is the magnitude of $\vec g$.
The equation of motion is now $m \dot{v}_Y \hat Y= -m\,g\hat Y - bv_Y \hat Y\Rightarrow m \dot{v}_Y = -m\,g - bv_Y.$

Now let's look at the terminal velocity with down as positive; $\vec v_{ter} =\dfrac{m\vec g}{b} \Rightarrow v_{ter} \hat y = \dfrac {mg}{b} \hat y\Rightarrow v_{ter} = \dfrac {mg}{b}$ where the magnitude of the terminal velocity is $\dfrac {mg}{b}$.

This leads to $y(t) \hat y = v_{ter}\hat yt + (v_{yo}\hat y - v_{ter}\hat y)\tau(1-e^{-\frac{t}{\tau}})\Rightarrow y(t) = v_{ter}t + (v_{yo} - v_{ter})\tau(1-e^{-\frac{t}{\tau}})$ where $y(t)$ and $v_{yo}$ are components and $v_{ter}$ is a magnitude.

With up as positive $\vec v_{ter} =\dfrac{m\vec g}{b} \Rightarrow \vec v_{ter} =\dfrac{mg(-\hat Y) }{b} = -\dfrac{mg}{b}(\hat Y)= -v_{ter} \hat Y$

This leads to $Y(t) \hat Y = \vec v_{ter}t + (v_{Yo}\hat Y - \vec v_{ter})\tau(1-e^{-\frac{t}{\tau}})\Rightarrow Y(t) \hat Y = - v_{ter}\hat Yt + (v_{Yo}\hat Y + v_{ter}\hat Y)\tau(1-e^{-\frac{t}{\tau}})$

$\Rightarrow Y(t) = -v_{ter}t + (v_{Yo} + v_{ter})\tau(1-e^{-\frac{t}{\tau}})$ where $Y(t)$ and $v_{Yo}$ are components and $v_{ter}$ is a magnitude.

  • $\begingroup$ This is a great explanation, thank you! $\endgroup$
    – Andrew
    Apr 9, 2020 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.