5. In $\triangle A B C$, $\angle A=60^{\circ}, A C=16, S_{\triangle A B C}=220 \sqrt{3}, B C=$
5. 49 .
$$
\begin{array}{c}
220 \sqrt{3}=\frac{1}{2} b c \sin A=\frac{1}{2} \times 8 c \sin 60^{\circ} \Rightarrow c=110, \\
B C=a=\sqrt{b^{2}+c^{2}-2 b c \cos A}=\sqrt{16^{2}+110^{2}-2 \times 16 \times 110 \times \frac{1}{2}}=49 .
\end{array}
$$