# Vertical motion with linear drag axis problem

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?

• Please provide information about the textbook which you are using. – Farcher Apr 8 '20 at 22:57
• John Taylor, "Classical Mechanics" – Andrew Apr 8 '20 at 23:16

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.

• This is a great explanation, thank you! – Andrew Apr 9 '20 at 11:13