What does it mean that a Lagrangian doesn't depend on time? What does the motion look like?

In fact for most of our examples, the Lagrangian will not depend explicitly on time: this means that $t$ does not show up in the expression for $\mathcal{L}=T-U$. For example, for a projectile in a uniform gravitational field, $\mathcal{L}=T-U = \frac{1}{2} m \dot{x}^2+ \frac{1}{2} m \dot{y}^2 -mgy$. Time doesn't show up anywhere, so the Lagrangian is time-independent and $\frac{\partial \mathcal{L}}{\partial t} = 0$. Another way of saying this is that the Lagrangian is invariant with respect to time translation: if you add one day to the all the times in the problem, you'll get the same Lagrangian and the same equation of motion and the same behavior of the projectile. It's consistent with our experience that a projectile behaves the same on any day you choose!

A Lagrangian would not be time-invariant (i.e., would depend on time) if there were some net energy flow into or out of the system (i.e., work done on or by the system). In such a case $T$ or $U$ or both would depend on time, corresponding to energy change of the system, energy would not be conserved. This would not be a ``closed system.''