Apple falling from boat mast

Question

To better understand how gravity and movement works, a common question is to ask where an apple would land if dropped from the top of a mast on a moving boat (let's assume a boat with a motor but with a mast, and no wind blowing apart from the relative speed of the boat, and a constant speed, let's say 50km/h, it's a fast boat).

People often mistakenly answer it would land on the rear of the boat, because the boat is still advancing forward while the apple is falling straight down.

So it is explained that there is a conservation of movement, and the apple was moving at the same speed as the boat, and hence it would land at the base of the mast, and it is often added that it's the same case as if you jump on a moving train, you don't end up at the rear of the train, but instead land where you were before.

However it seems to me that those 2 examples are not the same. On a train, the compartment is closed, so the air is trapped and also moves with the train. With the boat case, there is no close environment, so as soon as it gets dropped, it loses contact with the boat and stops being accelerated by the boat motors (that counteract the water/air slowing it down, so that it maintains a constant speed), so starts being slowed down by the air around it.

Assuming a 20m mast, I think it takes roughly 2s for the apple to reach the boat, and I'd say that in 2s, the air has ample time to slow the apple enough so that it doesn't fall on the bottom of the mast, but rather closer to the rear of the boat. Am I wrong? How far would the apple land (if there's not enough numerical values, feel free to take any reasonable choice)?

Answer 1

You are not wrong of course. The examples are supposed to be idealized examples where we ignore air friction.

Answer 2

Assuming a 20m mast, I think it takes roughly 2s for the apple to reach the boat, and I'd say that in 2s, the air has ample time to slow the apple enough so that it doesn't fall on the bottom of the mast, but rather closer to the rear of the boat. Am I wrong? How far would the apple land (if there's not enough numerical values, feel free to take any reasonable choice)

What really matters is the force applied to the apple by the perceived wind. Your question has a hidden assumption that the boat is travelling in an area with a wind speed of 0 relative to the water, so the apple experiences a wind equivalent to the speed of the boat in the opposite direction of it's movement. There might actually be a wind to either side that would move the apple sideways or even a strong tailwind that would make the apple land closer to the FRONT of the ship.

Using a drag calculator and using 4 inches for the size of an apple, approximating cross-section as a circle, using a wind speed of 30mph, and using google's result of 0.36 for an apple's drag coefficient I get 0.3241N for the force of the wind on the apple. Using 5oz for the mass of the apple and another calculator I get an acceleration of the apple of 2.3 m/s. But after one second that will lower the apple's speed relative to the wind by about 5mph and therefore the force, and the apple will then be accelerated by the wind at about 1.6 m/s.

So given the wind is moving at 30mph relative to the boat, the apple will land about 3.5 meters from the point on the boat it would fall on naturally if the boat were moving at the same speed as the wind relative to the water and the air seemed calm on the deck.

Note that this is highly dependent on speed. Back in the 1800s a more common speed would be about 10mph and the calculation shows that would only move the apple about 0.5 meters before it hit the deck. If the ship had a 20mph tailwind that would be 0.5 meters in front of the point where it was dropped.

Answer 3

However it seems to me that those 2 examples are not the same. On a train, the compartment is closed, so the air is trapped and also moves with the train. With the boat case, there is no close environment, so as soon as it gets dropped, it loses contact with the boat and stops being accelerated by the boat motors (that counteract the water/air slowing it down, so that it maintains a constant speed), so starts being slowed down by the air around it.

As other answers have already pointed out, in physics 101 we are discussing an ideal situation, where the air drag is neglected. Once it is taken into account, the two situations become different, for the reasons described in the Q. This is why people often err in answering the boat question - because the answer based on no air drag contradicts our experience.

This situation could be modeled, assuming for simplicity drag described by Stoke's law: $$ m\frac{d\mathbf{v}(t)}{dt}=-\gamma\mathbf{v}(t) +m\mathbf{g}, $$ which in projections on the horizontal and vertical axes gives us: $$ m\frac{dv_x(t)}{dt}=-\gamma v_x(t),\\ m\frac{dv_y(t)}{dt}=-\gamma v_y(t) - mg $$ These are linear differential equations that are easy to solve (things would be a bit more difficult with more realistic quadratic drag.)

Related: Why do rain drops fall with a constant velocity?