5. Place eight balls with a radius of 1 in two layers inside a cylinder, such that each ball is tangent to its four adjacent balls, and is also tangent to one base and the side of the cylinder. Then the height of this cylinder is $\qquad$ .
5. $\sqrt[4]{8}+2$ As shown in the figure, the centers of the four balls in the upper and lower layers, $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ and $A$, $B, C, D$, are the vertices of two squares with side lengths of 2, and the circumcircles $\odot O^{\prime}$ and $\odot O$ of these squares form the top and bottom bases of a cylinder.
In $\triangle A^{\prime} A B$, $A^{\prime} A=A^{\prime} B=A B=2$.
Let the midpoint of $A B$ be $N$, and the projection of $A^{\prime}$ on the lower base be $M$,
then $A^{\prime} N=\sqrt{3}$, and points $O, N, M$ are collinear.
Also, $O M=O^{\prime} A^{\prime}=\sqrt{2}, O N=1$.
Therefore, $M N=\sqrt{2}-1, A^{\prime} M=\sqrt{\left(A^{\prime} N\right)^{2}-(M N)^{2}}=\sqrt{2 \sqrt{2}}=\sqrt[4]{8}$.
Thus, the height of the cylinder is $\sqrt[4]{8}+2$.