Equilibrium only means that net external force $\sum \vec{F}_{ext} = 0$ on the system. This means that the velocity $\vec{v}$ does not change with time, because there is no acceleration. This does not mean that $\vec{v} = 0$ though. So your statement that $v_0=0$ based only on force balance in the $\hat{z}$ direction is wrong.
See steady flow.
Edit: You should revisit some good mechanics texts and revise Newton's laws. $$\vec{F} = m\vec{a}, \: \vec{a} = \frac{d\vec{v}}{dt}$$
$$\vec{F} = 0 \Rightarrow \vec{a} = 0 \Rightarrow \frac{d\vec{v}}{dt} = 0 \Rightarrow \vec{v} = const$$ (constant in time). This is Newton's first law. If gravitational force and pressure gradient are your only forces, your velocity cannot change in force equilibrium ($f_g + f_p = \rho\frac{d\vec{v}}{dt} = 0$). That's why I said $v_0=0$ based only on force balance is wrong. It should be $v_0=const$. If you have initial conditions/ boundary conditions, then that's different.
In reality, we have drag forces which are usually $F_d \propto v^n (n>0)$, so your net force changes over time, which kills your velocity. Consider a tank of water moving up at $v_0 = const\neq 0$. Would you say that the water is not in equilibrium (i.e. $\vec{F}_{ext} \neq 0$ so $\vec{a} \neq 0$) because the tank is moving? Do you see the mathematical inconsistency here? You only calculate equilibrium based on forces and forces don't change between constant velocity frames because they are inertial frames (1), (2).
If you get this, you'll see that $u_0, v_0 = const$ in your original question.
