Quadratic resistance
Preparation - re-read the MathPhys topic [Separation of variables].
Air has a low viscosity compared to liquids. So, things like bicycles and baseballs at typical speeds are already in a regime where the quadratic air resistance is much more important than the linear term. We'll start by considering what things look like with only the quadratic term (no linear drag force):
$$\myv{f}(v) \approx -cv^2 \uv{v}$$
The equation of motion for our particle is then $$m \ddot{\myv{r}} = m \myv{g} - c v^2 \myv{v}$$
As in the case of linear air resistance, we can cast this as a differential equation for $v$ using $\ddot{\myv{r}}=\dot{\myv{v}}$: $$\begineq m \dot{\myv{v}} =& m \myv{g} -c v^2 \uv v\\ =&m \myv{g} -c v \myv v.\endeq$$
The $x$ component of this vector equation is: $$m\ddot v_x=-c\sqrt{v_x^2+v_y^2}v_x,$$ and the $y$ component is: $$m\ddot v_y=-c\sqrt{v_x^2+v_y^2}v_y.$$ Yuck! These are inextricably coupled together, and we should be forced to solve this numerically.
To illustrate some of the features of quadratic motion, we will instead consider some special cases:
Horizontal motion alone
If an object were moving horizontally (no gravity) in the horizontal direction, the equation of motion would be: $$m\dot v= m \frac{dv}{dt} = -c v^2$$
Using separation of variables, we arrive at this equation for the differentials: $$m \frac{dv}{v^2} = -c\,dt$$
Integrating both sides of the equation $$\begineq m \int_{v_0}^v \frac{dv'}{v'^2} &=& -c \int_0^t dt'\\ m\left[-\frac{1}{v'}\right]_{v_0}^v = m\left( \frac{1}{v_0} - \frac{1}{v}\right) &=& -ct\endeq$$
Solving for the velocity with $\tau = m/(cv_0)$: $$v(t) = \frac{v_0}{1+t/\tau}.$$
A plot of this $v(t)$ is shown (blue) with the the same initial velocity for the case of linear resistance (gray)--the same values of $v_0$ and $\tau$.
Position
Integrating $v(t)$ to get the position (taking the integration constant $x_0=0$):
The position is shown (with only the quadratic resistance (blue)) along with the the position for the case of linear resistance (gray). The striking thing is the lack of any obvious maximum distance travelled.
What do you expect *realistically*?
Vertical motion alone
Taking $y \gt 0$, the the gravitational force is $+mg$, and the resistance is $-cv^2$ (where $v(t)=\dot y(t)$). $$m \dot{v} = mg - cv^2$$
Terminal velocity $v_{ter}$ corresponds to $m \dot{v} = 0$ which, from the equation above happens at: $$v_{\text{ter}} = \sqrt{\frac{mg}{c}}$$
For a baseball in air, this works out to something like 80 mph.
Where should you go if you want to go really, really fast, with only gravity to accelerate you? Here, with data
Vertical velocity
Now the equation of motion can be written with $v_{\text{ter}}$ as $$\dot{v} = g\left(1-\frac{v^2}{v_{\text{ter}}^2}\right)$$
Separating those variables: $$\frac{dv}{1-(v^2/v_{\text{ter}}^2)} = g\,dt $$
Integrating (e.g. CoCalc or integrals.wolfram.com) : $\frac{v_{\text{ter}}}{g} \arctanh (\frac{v}{v_{\text{ter}}}) = t$
Rearranging a bit for $v(t)$ $$v(t) = v_{\text{ter}} \tanh\left(\frac{gt}{v_{\text{ter}}}\right)$$ Here is $v(t)$ vs $t$
Integrating once more and taking $y_0= 0$ $$y=\frac{v_{\text{ter}}^2}{g} \ln \cosh\left(\frac{gt}{v_{\text{ter}}}\right)$$ $y(t)$ vs $t$ below: