Vertical projectile motion with quadratic drag (sign convention)

[f3sicA_A]

Homework Statement

A projectile that is subject to quadratic air resistance is thrown vertically up with an initial speed $v_0$. (a) Write down the equation of motion for the upwards motion and solve it to give $v$ as a function of $t$. (b) Show that the time taken to reach the top of the trajectory is:

$$t_\mathrm{top}=(v_\mathrm{ter}/g)\tanh^{-1}{(v_0/v_\mathrm{ter})}$$

Relevant Equations

$$\pmb{f}=-cv^2\pmb{\hat{v}}$$

I am attempting problem number 2.38 from John R. Taylor's Classical Mechanics and I am not getting the correct answer. My procedure is as follows:

Equation of motion (taking up as the positive direction):

$$m\dot{v}=-mg-cv^2$$

Now to find $v_\mathrm{ter}$, the terminal velocity, we consider the downward motion of the ball, that is, the velocity of the ball is in the downward direction, and therefore, we get:

$$-mg+cv^2=0$$

$$\implies v_\mathrm{ter}=\sqrt{\frac{mg}{c}}$$

I have a feeling I have made some sign convention error here but I am not sure what is wrong in my understanding. Continuing with this chain of reasoning:

$$\dot{v}=-g\left(1-\frac{v^2}{v_\mathrm{ter}}\right)$$

$$\implies \int_{v_0}^{v}\frac{1}{1-v'^2/v_\mathrm{ter}^2}\,\mathrm{d}v'=-g\int_0^t\,\mathrm{d}t'$$

$$\implies v_\mathrm{ter}\left[\tanh^{-1}{\frac{v'}{v_\mathrm{ter}}}\right]_{v_0}^v=-gt$$

$$\implies v_\mathrm{ter}\tanh^{-1}{\frac{v}{v_\mathrm{ter}}}=-gt+v_\mathrm{ter}\tanh^{-1}{\frac{v_0}{v_\mathrm{ter}}}$$

$$\implies v=v_\mathrm{ter}\tanh{\left[-\frac{gt}{v_\mathrm{ter}}+\tanh^{-1}{\left(\frac{v_0}{v_\mathrm{ter}}\right)}\right]}$$

From this, if I substitute $v=0$ to find $t_\mathrm{top}$, I get:

$$t_\mathrm{top}=\left(\frac{v_\mathrm{ter}}{g}\right)\tanh^{-1}{\left(\frac{v_0}{v_\mathrm{ter}}\right)}$$

Whereas the answer requires that the function is arctan instead of hyperbolic arctan. Please let me know where I am going wrong, thank you!

Note: I am new to the forum and I am not very well-versed with the rules of this platform (I went through the basic guidelines) so in case I make some mistakes, please overlook them (and do let me know if I can do something to improve my presence on this platform). For context, I am an undergraduate student self-studying classical mechanics in the summer before my semester begins, thank you!

 

[kuruman]

You have $v^2_{\text{ter}}=\dfrac{mg}{c}\implies c=\dfrac{mg}{ v^2_{\text{ter}} }.$

Replace that in the original equation. The relative sign between the two terms on the right hand side in the original equation is positive. Your algebra somehow changed it to negative. Redo the algebra more carefully this time.

It looks like this LaTeX does not recognize \arctanh ($\arctanh$) but it recognizes \tanh^{-1} ($\tanh^{-1}$).

 

[f3sicA_A]

The relative sign between the two terms on the right hand side in the original equation is positive. Your algebra somehow changed it to negative. Redo the algebra more carefully this time.

Thank you for your response! I am not sure which equations you are referring to, but I am guessing, I have made a mistake in the following equations:

$$m\dot{v}=-mg-cv^2$$

$$-mg+cv^2=0$$

I realize that I have changed the signs here; however, my logic behind this is that the in the first equation, the magnitude and direction $-cv^2$ is applied on the unit vector $\pmb{\hat{v}}$; however, to take into account the changing direction of the unit vector itself, in the second equation where I write $+cv^2$, it is as a magnitude of the unit vector $\pmb{\hat{x}}$, and for doing this, I multiply the original magnitude $-cv^2$ by $\cos{\theta}$ where $\theta$ is the angle between $\pmb{\hat{v}}$ and $\pmb{\hat{x}}$. Is this line of thinking faulty, or are you referring to some other equations?

Also, thank you for informing me about the LaTeX command for inverse hyperbolic tangent function, I have corrected that in the original question, along with the inline formulae.

 

[hutchphd]

Terminal velocity happens when $\dot v=0$. Plug that into your first equation. Stop. Do not wave your hands. Why are you trying to wave your hands???
In 1 dimension a signed number is a vector

 

[kuruman]

I am not sure which equations you are referring to $\dots$

$m\dot{v}=-mg-cv^2$

$v^2_{\text{ter}}=\dfrac{mg}{c}\implies c=\dfrac{mg}{ v^2_{\text{ter}} }.$

$m\dot{v}=-mg-\dfrac{mg}{ v^2_{\text{ter}} }v^2$

$\dot {v}=~?$

 

[kuruman]

In 1 dimension a signed number is a vector

Normally yes. In situations of this sort where you have damping that always opposes the velocity, I find that considerable confusion and grief are avoided if one follows the rule "choose as positive the direction of motion." Then the velocity and the speed are the same and one writes

$m\dfrac{dv}{dt}=-bv-mg~~$ if the mass is moving up and both the damping force and the weight reduce the speed

$m\dfrac{dv}{dt}=-bv+mg~~$ if the mass is moving down and the damping force reduces the speed while the weight increases it.

In either case symbol $v$ stands for speed not velocity which makes what's going on transparent.

 

[f3sicA_A]

$m\dot{v}=-mg-cv^2$

$v^2_{\text{ter}}=\dfrac{mg}{c}\implies c=\dfrac{mg}{ v^2_{\text{ter}} }.$

$m\dot{v}=-mg-\dfrac{mg}{ v^2_{\text{ter}} }v^2$

$\dot {v}=~?$

I realized where I made the mistake in signs, and now find it quite silly. I thank you for pointing that out.