# Sign problem in projectile with drag

So recently, I've been playing around with the variables in projectile motion with drag to see how the trajectory changes. I had my ODE and solved it using separation of variables but the equation didn't really work as well as I thought.

Here's the ODE

$$m\frac{\mathrm dv}{\mathrm dt}=bv^2-mg$$

After solving it by substituting

$$v=\alpha\cdot \tanh(u)$$

After I got the solution and solved for the initial conditions, I got to a solution which worked as long as $v>0$. I realized that in the ODE, instantaneous velocity is squared so it loses its sense of direction causing the solution to go out of wack.

Is it possible to fix this issue and be able to solve the equation analytically?

I was thinking to change the ODE to

$$m\frac{\mathrm dv}{\mathrm dt}=bv\cdot \operatorname{sign}(v)-mg$$

I guess I can also solve it by solving for two ODE, but I really really just wanted one elegant solution (not piecewise), but I'm having difficulty solving it.

• Don't quite understand what you're doing. Initially you need to have $b v^2 > mg$ otherwise the rocket will never get off the launching pad, right? And if this condition is satisfied then v will continue to monotonically increase with time so you never have to worry about it becoming negative. – Samuel Weir Aug 30 '18 at 3:00
• It looks like this question is actually not about the rocket equation, instead it concerns the formula for a falling object with (idealized) air resistance. Nick, is that correct or am I missing something? I think it would help if you edited to make that consistent. Also @SamuelWeir that comment looks like it should be an answer – David Z Aug 30 '18 at 3:52
• @DavidZ - I put it in as a comment asking for clarification because at this stage I'm still not sure what physical problem the OP is trying to answer here or if the equation he wrote down really expresses what he wants it to express. – Samuel Weir Aug 30 '18 at 4:13
• @SamuelWeir Ah, yeah I suppose you're right. The latter part of it looks kind of answer-ish but that's all. – David Z Aug 30 '18 at 5:07
• The eq. looks more like a projectile with quadratic drag than a self-propelled rocket. – Qmechanic Aug 30 '18 at 5:30

OP is correct, in the case of quadratic resistance as opposed to linear resistance, the upwards and downwards segments are typically split up into two separate EoMs to accommodate the change in sign of $v$. See e.g. P.541 Section III in ref$^1$ for comments.
The generic equation modelling the whole trajectory is your last displayed equation indeed but to proceed analytically one must consider the cases $\text{sign}(v) = \pm 1$ separately, analogously as to how one would deal with equations involving the modulus function, $|x|$, for example. In one case, you will have a function governed by a $\tan$ and in the other a $\tanh$. You could then concoct a single solution through the use of theta functions if you so wish.
$$v(t)=\tanh \left( \left( -t\,c+m\operatorname{arctanh} \left( {\frac {{\it v0}\,b}{c} } \right) \right) {m}^{-1} \right) c\,{b}^{-1}$$
with $c=\sqrt{b\,m\,g}$ and $v(0)=v0$
This solution is defined also for $v < 0$ but you have to take care about the $\tanh$ function.