At point $O$ on the shore of the Dongjiang Lake (the lake shore can be considered a straight line), a rescue boat is parked. Due to the sudden breakage of the rope, the boat is blown away, with its direction forming a $15^{\circ}$ angle with the shore, at a speed of $2.5 \mathrm{~km} / \mathrm{h}$. At the same time, a person starts chasing the boat from the same location on the shore. It is known that his speed on the shore is $4 \mathrm{~km} / \mathrm{h}$, and his swimming speed in the water is $2 \mathrm{~km} / \mathrm{h}$. Can this person catch up with the boat? If the boat's speed changes, what is the maximum speed of the boat that can still be caught by this person?
Let's analyze the given problem. Suppose this person runs to point $A$ on the shore and then enters the water, catching up with the boat at point $B$. Let the boat's speed be $v$, the time it takes for the person to catch up with the boat be $t$, and the time the person spends chasing the boat on the shore be $k$ times $t$ $(0 \leqslant k < 1)$.
\begin{aligned}
& f(k) = 12k^2 - 2(\sqrt{6} + \sqrt{2})vk + v^2 - 4.
\end{aligned}
(1) If $f(0) \leqslant 0$, then there must exist $k_{0} \in [0,1)$ such that $f(k_{0}) = 0$.

In this case, $f(0) = v^2 - 4 \leqslant 0$, solving this gives $v \leqslant 2$.
(2) If $f(0) > 0$, to ensure that $f(k)$ has a solution in $[0,1)$, then

we need to solve
\[
\left\{\begin{array}{l}
\Delta = [2(\sqrt{6} + \sqrt{2}) v - 8]^2 - 4 \times 12(v^2 - 4) \geqslant 0, \\
\frac{2(\sqrt{6} + \sqrt{2}) v - 8}{2 \times 12} \in (0,1).
\end{array}\right.
\]

Thus, $2 < v \leqslant 2 \sqrt{2}$.
In summary, when $v \leqslant 2 \sqrt{2}$, the person can catch up with the boat.
Therefore, when the boat's speed is $2.5 \mathrm{~km} / \mathrm{h}$, the person can catch up with the boat, and the maximum speed at which the boat can be caught is $2 \sqrt{2} \mathrm{~km} / \mathrm{h}$.