when there is air resistance the horizontal component of velocity is not constant anymore and it decays.
Let's set the origin at the launch point with the X, Z axes, and C as positive constant friction or drag of the air.
The equation of motion:
$$
m\,\frac{d{\bf v}}{dt} = m\,{\bf g} - c\,{\bf v},$$
$$
V = ( V_X, V_Z) \quad g=(0,-g),v_t =terminla \ velocity=mg/c$$
$$ \frac{dv_x}{dt}= - g\,\frac{v_x}{v_t},$$
$$ \frac{d v_z}{dt}= - g\left(1+\frac{v_z}{v_t}\right).$$
Integrating X component
$$
\int_{v_{x\,0}}^v\frac{dv_x}{v_x} = - \frac{g}{v_t}\,t,
$$
Wher $ v_{x\,0} = v_0\,\cos\theta$
is the $x$-component of the launch velocity. therefore
$$\ln\left( \frac{V_X}{V_{X_0}}\right)= \frac{-g}{V_t}t, \ then $$
$$V_x=v_0cos\theta*e^{-gt/v_t} $$.
The x component will decay exponentially.
Applying the vertical forces and integrating we get:
$$V_Z=V_0sin\theta*e^{-gt/V_t}-V_t* \left(1-e^{-gt/V_T} \right)$$
Among other things, the above implies that if the ball stays in the air much longer than the order of $,V_t/g \ $it will fall vertically.
I know this answer can use some details and graphs, I come back and finish them.