19. (15 points) In mid-August 2006, the city of Zixing in Hunan Province encountered a once-in-a-century flood disaster. At point $O$ on the shore of Dongjiang Lake (consider the lake shore as a straight line), a rescue boat was parked. Due to the sudden breakage of the rope, the boat was blown away, with a direction making a $15^{\circ}$ angle with the shore, at a speed of $2.5 \mathrm{~km} / \mathrm{h}$. At the same time, a person on the shore started chasing the boat from the same location. It is known that his speed on the shore is $4 \mathrm{~km} / \mathrm{h}$, and his 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 this person can catch up with?
19. As shown in Figure 3, 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 $t$ $(0 \leqslant k < 1)$.
$$
(1) \text{ If } f(0) \leqslant 0, \text{ then there must exist } k_{0} \in [0,1), \text{ such that }
$$
$$
f\left(k_{0}\right)=0.
$$

In this case, $f(0)=v^{2}-4 \leqslant 0$, solving this gives $v \leqslant 2$.
(2) If $f(0)>0$, for $f(k)$ to have a solution in $[0,1)$, then
$$
\left\{\begin{array}{l}
\Delta=[2(\sqrt{6}+\sqrt{2}) v-8]^{2}-4 \times 12\left(v^{2}-4\right) \geqslant 0, \\
\frac{2(\sqrt{6}+\sqrt{2}) v-8}{2 \times 12} \in (0,1).
\end{array}\right.
$$

Solving this gives $\left\{\begin{array}{l}v \geqslant 2(\sqrt{6}+\sqrt{2}) \text{ or } v \leqslant 2 \sqrt{2}, \\ \sqrt{6}-\sqrt{2}<v<4(\sqrt{6}-\sqrt{2}).\end{array}\right.$
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 $2 \sqrt{2} \mathrm{~km} / \mathrm{h}$.
(Ouyang Xinlong provided)