How does 5 mph wind catch a 10 mph boat? My original posted question and the ans

Question

How does 5 mph wind catch a 10 mph boat?

My original posted question and the answer provided (which I believe is 100% correct, I just can't explain why another possible answer is wrong - hence this question):

If a boat is drifting down a river with the current at 10 mph, and a 5 mph wind is blowing from directly behind the boat (and it has full pirate sails), then what will be the velocity of the boat?

10 mph, 15 mph, or some other value? Ignore friction, ballast, quality of seamanship of the boat's captain, etc.

So, the new question: If you were standing on a fixed platform in the middle of the river as the boat went by, and you threw a rock at 5 mph at the boat, the rock would fall in the water rather than hitting the boat because of their relative velocities. Why is the same not true of a 5 mph wind? By that logic, V(boat) would be 10 mph, because the wind never "catches" the boat. The formula would be : V(boat) = Max [ V(wind), V(river)] , rather than the sum. This is not correct ... but why not?

Thanks!

Explanation / Answer

Wind is blowing continuously from behind the boat, it not starts after the boat, wind is blowing before the boat starts. It is naturally, wind is not a particle like stone. Wind covers whole the environment in which boat is moving. Wind put the pressure on the boat when boat moves and speed of boat increases.

But stone is like a particle. Stone can't cover the environment around the boat.

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