6. As shown in Figure 1, given that line $l \perp$ plane $\alpha$, with the foot of the perpendicular being $O$, in the right triangle $\triangle ABC$, $BC=1, AC=2, AB=\sqrt{5}$. The right triangle moves freely in space under the following conditions: (1) $A \in l$, (2) $C \in \alpha$. Then the maximum distance between points $B$ and $O$ is
6. $\sqrt{2}+1$.

It is easy to know that when the distance between points $B$ and $O$ is maximized, points $A$, $B$, $C$, and $O$ are coplanar.
Let $\angle A C O=\theta$. Then $O C=2 \cos \theta$.
In $\triangle B C O$,
$$
\begin{array}{l}
|O B|^{2}=|O C|^{2}+|B C|^{2}-2|O C||B C| \cos \left(90^{\circ}+\theta\right) \\
=4 \cos ^{2} \theta+1+4 \sin \theta \cdot \cos \theta \\
=2 \cos 2 \theta+2 \sin 2 \theta+3 \\
\leqslant 3+2 \sqrt{2} .
\end{array}
$$

When $\theta=\frac{\pi}{8}$, the equality holds.
Thus, $|B O|_{\max }=\sqrt{2}+1$.