I way to consider this problem is without analysing the exact maths, for whether the ball is dropped from .3m or .5m, but look at the problem in terms of its limits.
It should be easy to see that if the ball is dropped from the same height as the mans feet (i.e. not dropped at all) it is accepted that the ball will start and end precisely in the middle of the mans feet.
However, at the opposite end, if the ball starts at the maximum height (i.e. centre of the cylinder) it will not fall at all, because it will have no scalar veloicity component at all. In this case, the man would continue to rotate for an infinite number of rotations, before the ball lands.
By moving the ball very slightly from this centre point, so only a very slight scalar velocity, the mans number of rotations would still be very high before the ball lands. This limit can show that the ball is conceptually to the right of the mans feet.
As the position of the ball is lowered, the number of rotations the man completes will decrease, but it will always be landing to the right of the mans feet until the initial limit is reached.
Hope this helps
Gavin

I just started on the problem, but I found a couple of complications. First even though the person's feet and the ball have the same angular velocity their linear velocity is different, the feet are actually moving faster than the ball. The second is that the ball's rate of acceleration is not constant. Another way to look at it is the weight of the ball increases as it falls, as apposed to its mass which remains constant. In fact if the ball is held at the center point it would be weightless and not fall at all. I think this means the ball would always land to the left of center.

I don't agree. Since when the ball is being held, the only force the person is applying to it a a force which points always towards the center of the cylinder, I have a hard time thinking that the ball will end up anywhere other than right between the person's feet when that force is removed. Since I'm at work, I can't really follow this through formally, and I know gut feelings are not he same as physics, but this doesn't feel right.

Wouldn't centrifugal force act on the ball perpendicular to the direction of movement? I don't recall how to predict the path in that case (was it but certainly the path will be more complicated than previously described.

Randa11 is correct. But while rotation speed doesn't matter, the height of the ball does. For example, if the ball is dropped from about 8.71625 meters, it will land directly between his feet (after the station has completed a full rotation). The formula for distance between the ball and the center of his feet is:
r = radius of station
h = height of ball when dropped
d = distance from center of feet to where ball will land (to his left, assuming he is facing us)
d=[r/(r-h)*sqrt(2*r*h-h^2)]-[r*arctan(sqrt(2*r*h-h^2)/(r-h))]

The trick is to notice that the mans belly button (or whatever) and the ball continue to move at the same scalar speed, and the bottom of the triangle will always be greater than the arc traced by the belly button to the same angle.
Let the angle the ball is at when it hits the floor be ß
and, call the slightly larger angle traced out by the belly button ß+
Clearly the ball hits the floor after it has travelled d=sqrt(19) (pythagoras)
and so, by that time the mans belly button has gone the same distance so (ß+)=sqrt(19)/9
the ball hits the floor at distance b=10*(ß)
and the floor or mans feet have moved f=10*(ß+)
It doesn't matter how quickly the space station is rotating.
ß =25.84193ş
ß+=27.74961ş
b =4.510268 m
f =4.843221 m
distance from centre of mans feet = f - b = 0.332953 m

Making the Grade in Industrial Design Rich Quinnell16 comments As every developer knows, there are the paper specifications for a product design, and then there are the real requirements. The paper specs are dry, bland, and rigidly numeric, making ...

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