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Think about what happens in the limit that the release point
approaches the center of the station. In that limit, the ball
never hits the "ground." Now, unless you think that limit is
approached discontinuously, the implication is that the ball must
fall further and further behind the astronaut's feet as the height
of the release point increases.
Indeed, good explanations of this fact exist for both the inertial
and rotating observer. In the inertial frame, the distance
traveled by the ball is sqrt(2Rh+h^2) where h is the height of the
release and R is the radius of the space station and the speed of
the ball is omega*(R-h) so the time it takes the ball to fall is
srqt[ 2*(h/R) + (h/R)^2 ]
t_b = -------------------------
omega*(1 - h/R)
(Notice what happens as h/R --> 1.)
To reach the place where the ball will fall, the astronaut's feet
must move through an angular displacement of arccos(1 - h/R), so
the time it takes the feet to get there is
arccos(1 - h/R)
t_f = ---------------
omega
Now it is not *too* difficult to show that t_f < t_b for all h >
0. Perhaps the easiest way to see this is simply to plot
srqt[ 2*(h/R) + (h/R)^2 ]
t_b/t_f = -------------------------
(1 - h/R)*arccos(1 - h/R)
as a function of h/R and note that it is a monotonically
increasing function. Thus, the ball lands behind the feet.