Hyperbolic metric geodesically complete

Consider the upper half plane model of the hyperbolic space ($\mathbb{H}$ with the riemannian metric $g=\frac{dx^2+dy^2}{y^2}$). It is known that $(\mathbb{H},g)$ is geodesically complete, which means that no geodesic can reach the border $\partial \mathbb{H}$ in a finite time.

Why is that?

Of course if I consider particular geodesics such as $t\mapsto (0,e^{-t})$ this is true, but I can't figure out a general proof for this fact which doesn't rely on the particular form of the geodesic considered. My intuition is that it must depend on the fact that, as a geodesic approaches the border, the denominator of $g$ goes to $0$. Can you point me out an explanatory proof which shows that the length of a hyperbolic geodesic which goes to the border can't be finite?

EDIT: Thank you so much for the answers, I realized I didn't make myself clear. What I mean is: is there a way to prove directly for any hyperbolic geodesic (using the fact that $g$ has $y^2$ at the denominator) that its length is infinite?

Because $\mathbb{H}$ has many symmetries doing it for one geodesic is enough. You can use an appropriate Mobius transformation, $g$, to realize any geodesic $\gamma$ of $\mathbb{H}$ as

$\gamma(t)=g\cdot ie^{-t}$.

Edit: I saw in your question you didn't want to use explicit geodesics, so you could also do something like this (to see why the denominator forces infinite length)

Let $\gamma(t)=x(t)+iy(t)\subset\mathbb{H}$ be a curve with $\gamma(0)=x_0+iy_0\in\mathbb{H}$ and $\gamma(1)\in\partial \mathbb{H}$ (and say $\dot y<0$).

Then $\ell(\gamma)\ge \int_0^1\frac{|\dot y|}{y} dt=-\int_{y_0}^0 \frac{dy}{y}=\log y|_{0}^{y_0}=\infty$.

A bit silly, since as mentioned, you can prove it by hand, but: covers of complete manifolds are complete, and compact manifolds are automatically complete. There are a great deal of hyperbolic structures on $\Sigma_g$, $g>1$.