5. In tetrahedron $ABCD$, it is known that $AD \perp BC$,
\[
\begin{aligned}
AD=6, BC & =2, \text{ and } \\
AB+BD & =AC+CD=t(t \in[8,+\infty)) .
\end{aligned}
\]

Then the maximum value of $V_{\text{tetrahedron } BCD}$ is in the range \_\_\_\_\_\_
5. $[2 \sqrt{6},+\infty)$.

Draw $B F \perp A D$ at point $F$, and connect $C F$.
Since $A D \perp B C$, we know $A D \perp$ plane $B F C$.
Therefore, $C F \perp A D$.
Let $A B=x, A C=y$.
From $A B^{2}-B D^{2}=A F^{2}-D F^{2}=A C^{2}-C D^{2}$
$$
\Rightarrow x=y \Rightarrow B F=C F \text {. }
$$

Then $V=V_{\text {tetrahedron } A-B F C}+V_{\text {tetrahedron } D-B F C}$
$$
=\frac{1}{3} S_{\triangle B F C} \cdot A D=2 \sqrt{B F^{2}-1} \text {. }
$$

Notice that, $A B+B D=t>6$.
By the definition of an ellipse, we know that when $A B=D B=\frac{t}{2}$,
$$
B F_{\max }=\sqrt{\frac{t^{2}}{4}-9}, V_{\max }=2 \sqrt{\frac{t^{2}}{4}-10} \geqslant 2 \sqrt{6} \text {. }
$$