In physics education you often consider "real world problems" with projectile motion. Most times in introductory courses you neglect air drag. But how can students (knowing nothing about Reynolds numbers etc. at that point) see if it is really negligible in the current problem or not. One possible procedure would be as follows: Calculate the maximum speed $v_{\mathrm{max}}$ of the projectile in the problem by simply neglecting the air drag. Then calculate the max. air drag force, that would act at this speed $F_{\mathrm{Drag,max}} = k \cdot v_{\mathrm{max}}$ to get an upper bound for the actual maximal air drag force you would see if you solved the problem including air drag. Then compare $F_{\mathrm{Drag,max}}$ to the gravity force $F_{\mathrm{G}}$ (compare here, 3.2 for a similar but more crude approach), i.e. look at the quotient $q = \frac{F_{\mathrm{Drag,max}}}{F_{\mathrm{G}}}$. If $q$ is small you would conclude that you may neglect it. "Small" is mainly defined by the accuracy you want to solve the problem (keep in mind that if you measure something in physics you have measurement errors). So, for example you say, 1% accuracy is enough for the problem you might say, that the air drag is negligible if $q < 1\%$.
Edit:
What would be a reasonable to convert this idea into a mathematical theorem and how to prove it.
In particular one needs find a reasonable definition of what "negligible" means in this context in dependence of $q$. I suggested (see below) a way to express this mathematically, but as David K. pointed out in his answer this wouldn't work. Maybe one needs to include a timescale or space scale of the problem (defined by when or where the projectile hits the ground).
Old version of the rest of the question
Is this procedure correct from a mathematical point of view? I.e. let the projectile motion start at time $t = 0$ and end at time $t_{\mathrm{F}} > 0$ (that's where it hits the ground). Is it then true, that
$$\frac{|\vec{r}(t) - \vec{r}_{\mathrm{Drag}}(t)|}{|\vec{r}(t)|} \leq q$$ and
$$\frac{|\vec{r}(t) - \vec{r}_{\mathrm{Drag}}(t)|}{|\vec{r}_{\mathrm{Drag}}(t)|} \leq q$$ for all $t \in [0;t_{\mathrm{F}}]$, where $\vec{r}(t) \in \mathbb{R}^3$ denotes the position at time $t$ of the projectile if calculated without air drag, $\vec{r}_{\mathrm{Drag}}(t) \in \mathbb{R}^3$ the position at time $t$ if calculated including air drag and $|\cdot|$ denotes the euclidian norm. In other words: is it true that the position you get by neglecting air drag differs maximal by $100 q \,\%$.
To state the problem more mathematically consider the following ODE:
$$ m \cdot \ddot {\vec{r}_{k}} = m \cdot \vec{g} - k \cdot |\dot {\vec{r}_{k}}| \cdot \dot {\vec{r}_k} $$
With the condition that $\dot {\vec{r}_k}(0) = \vec{v}_0 \in \mathbb{R}^3$, $\vec{r}_k(0) = \vec{r}_{k,i} \in \mathbb{R}^3$, $k > 0$ (determines strength of the drag, $k = 0$, no drag) and $\vec{g} = (0,0,-g)$ with $g > 0$ as well as $m > 0$ (physically the mass of the projectile). Additionally assume that the third component of $\vec{r}_{k,i}$ is positive (i.e. that the particle starts above the ground) and that $t_{k,\mathrm{F}}$ is the smallest $t > 0$ where $\vec{r}_k(t) = 0$.
The solution $\vec{r}_k\colon [0;t_{k,\mathrm{F}}] \subset \mathbb{R} \to \mathbb{R}^3$ may be assumed to be $C^{\infty}$.
Then let $v_{0,\mathrm{max}} := \mathrm{max} \{|\dot {\vec{r}_0}(t)| \mid t \in [0;t_{\mathrm{0,F}}] \}$ and $q := \frac{k\cdot v_{0,\mathrm{max}}^2}{m\cdot g}$.
Is it true that $$ \frac{|\vec{r}_0(t) - \vec{r}_k(t)|}{|\vec{r}_j(t)|} \leq q $$ where $j \in \{0;k\}$ für all $t \in [0;\mathrm{min}\{t_{0,\mathrm{F}},t_{k,\mathrm{F}}\}]$.
If so, how to prove it?. If not, what would be an counter example and how could the proposition be modified to get a similar line of reasoning for the physicist as described above?
