Question: For instance, if I have encountered

\begin{equation}
\label{eqq2}
 \frac{\partial \mathcal{L}}{\partial (\partial\_i \phi)} = 0
\end{equation}

This tells us that $\mathcal{L}$ cannot depend on $\partial\_i \phi$, only on the field $\phi$ and it's time-derivative $\dot{\phi}$.

Does this go against some rule for field theories? If so, I can say ok, the Lagrangian must not only depend on the things I guessed it did in my ansatz.

Answer: 
I don't understand how you concluded that the ball is slowing down. It isn't. Its acceleration is decreasing. If it were slowing down, the velocity-time graph would curve downwards. Here it is flattening which shows that its acceleration is approaching $0$.

According to Newton's second law:

\begin{equation}
F\_{net} = m a
\end{equation}

When the ball is falling down, it is experiencing two forces; the weight of the ball itself and the drag force. The drag force is proportional to the velocity, so after attaining a specific velocity called the terminal velocity, the weight equals the drag force, $F\_{net}$ becomes $0$ and the ball stops accelerating, which results in the flattening of the velocity-time graph shown.

If the drag force were greater than the weight, the ball would experience a net upwards force and hence would keep on moving upwards instead of coming down.
For a sphere drag force = 6πnrv (stokes' law) {v=velocity} so in start v=0 so drag =0 so net force is downwards
so as ball comes down net force is downwards so velocity of increases in downward direction until for some v (drag force)=(gravity)
at this point net acceleration is 0 thus velocity is constant therefore drag force is constant and gravity is constant at every moment.
So from this point onward velocity is constant & known as terminal velocity.

So from starting only the velocity is increasing till terminal velocity and not that velocity first goes to maximum and then decreases