People can only catch up if they run along the coast for a while and then swim to catch up with the ship.

Therefore, only when the trajectory of people running along the coast, swimming in the water and sailing in the water form a closed triangle can people catch up with the ship.

Assuming that the ship speed is x (unknown),

It can be seen that when x≥4, it is impossible for people to catch up with the ship.

When 0 < x ≤ 2, people can catch up with the ship directly from the original groundwater without running ashore.

Therefore, 2 < x < 4.

It takes one person t to catch up with the ship, including kt (0 < k < 1) on shore and (1-k)t in water.

Distance traveled by boat: xt

Distance of shore personnel: 4kt

Distance of people in water: 2 (1-k) t.

The three forms a triangle, and the angle between the first two is 15.

By cosine theorem: (4kt) 2+(XT) 2-2 (4kt) (XT) cos15 = 4 (1-k) 2t2.

Cos 15 = (√ 6+√ 2)/4, substituted and sorted:

12k^2-[2(√6+√2)x-8]k+x^2-4=0

The equation about k has a real number solution in (0, 1).

Therefore, δ = [2 (√ 6+√ 2) x-8] 2-48 (x 2-4) ≥ 0.

And the product of two roots satisfies: 0 < (x 2-4)/ 12 < 1.

Solution, 2 < x ≤ 2 √ 2

That is, when the ship speed is within the range of (2,2 √ 2), people can catch up with the ship.

2 < 2.5 < 2 √ 2, so when the ship speed is 2.5km/h, people can catch up with the ship.

The maximum speed that the ship can be surpassed = 2 √ 2 km/h.