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What height is the ship, what height is the observer (eye height)? We need a few more parameters.

When we have those, we can calculate it as described here: https://physics.stackexchange.com/questions/593064/how-to-calculate-how-much-of-an-object-is-hidden-due-to-earth-curvature

Earth's radius is 3959mi or 6371km.

Or we use this simpler formula: h = 6,371km * (1 - cos(0.009° * d))
We could solve for d here if h is known (too tired for that right now).
Let's assume the observer is on water level, and the ship is 30m high.

6371 * (1 - cos(0,009 * 19,55)) = 30m
So that gives us roughly d = 19,55km fo ar distance with the assumption of no atmospheric effects going on for the ship to stay visible longer or shorter.

Good. We know really well how it works exactly, we understand the seasons and sunlight very well. But the good thing is that even if "NASA" (most science) was totally off, the scientific method would allow us to gradually gather new evidence until we have figured out a complete, working model.
"In an ideal world, flat Earthers would some day once again figure out that earth is in fact a spheroid" - unless they keep holding themselves back by biases.

This is what 24h Sun looks like on flat earth, this is the winter Sun lighting FE:


FE vs. GE

What do you think, which model explains it better?

Because the FE animation is what would have to happen on a flat surface, regardless of the map being a EAP map (globe projection).