In the main (but NOT QUITE) the wrench will fall directly below the sailor. However there's a tiny difference because one of the motions involved is NOT a uniform motion in a straight line... Read on, this is worth your time!

The reason why there is a small difference is the so-called Coriolis effect (see link) due to the rotation of the Earth. The fact that the ship is moving is utterly negligible here (sailboats don't go round the world in 24 hours) and we may as well assume, for the sake of simplicity, that the mast is firmly planted on the beach! Also for the sake of simplicity, we'll assume that the beach is on the equator (that's where the effect is largest anyway).

If you look at the scene (with a telescope) from above the North Pole, tou see the equator rotating counterclockwise (see second link). Let R be the radius of the Earth (sea-level) and h be the height of the mast.

In a nonrotating frame of reference, a wrench that's dropped from the top of the mast is really dropped with an horizontal velocity which equals the velocity of the tip of the mast, namely -A(R+h) if we call A the sidereal angular velocity of the Earth (2*pi radians in a sidereal day of 86164.09 seconds). I am using a minus sign just to be consistent with the traditional orientation of an (x,y) system where y is positive upward and x is positive to the left.

The key point is that the wrench maintains this horizontal velocity -A(R+h) throughout its fall but the ground moves at a lesser velocity (in absolute value) -AR. If there was no rotation, both velocities would be equal, but the rotation of the Earth makes them different. If it takes a time t for the wrench to hit the ground, it will therefore do so at a position -Aht with respect to the bottom of the mast. This is slightly TO THE EAST of the bottom of the mast. Let's be quantitative:

Since t (the time of fall) is sqrt(2h/g) where g is the acceleration of gravity the distance Aht in meters is proportional to h^(3/2) [the cube of the square root of h] with a coefficient of proportionality equal to (2*pi/86164.09)sqrt(2/9.8) or about 0.000033. For a tall 20m mast, this means a deviation to the east of about 0.003 m (3 mm).

3 mm is small but it's not negligible (it's more than one tenth of an inch). For the height of a skyscraper (300 m) the eastward distance would be 17 cm (almost 7 inches).

Now, in a slowly moving ship (traveling to the west, the east, the north or the south, it does not matter) the distance is still always 3 mm to the east.

Roughly speaking, this eastward distance of 3 mm is the distance traveled in the time of a 20 m fall (about 2 seconds)by a point on a circle of radius 20 m which rotates at a rate of one turn per day.

To estimate the Coriolis effect away from the equator and/or in more complicated circumstances (e.g., an artillery shell) would require more complicated mathematics, but the above does capture the essential... Enjoy.

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