Solution:

The motion of the ball on Earth looks like this:

The ball can go up 10 meters high on the Earth, so the initial velocity to upward direction is

Now we move on to the space station. Throwing the ball straight up while in a rotating space station will have an interesting effect. There is no gravity in space, so the ball will not follow a parabola, but will instead fly completely straight. From the point of view of the thrower though, the ball will appear to go up, and then return.
The reason that the ball will come back to the outer shell of the cylinder is because the initial velocity of the ball not only has the motion perpendicular to the edge of the shell, but the motion tangential to the shell as well. You must remember that the shell is rotating at a pretty fast rate to recreate a similar centripital force to Earth.
The motion of the ball in the space station looks like this (not drawn to scale, and remember there is no force acting on the ball while it is in the air, that is why it flies straight!):

We have to find the initial velocity both perpendicular and tangent to the surface of the shell.

First we find the angular velocity of the cylinder needed to equal g.

Now we find the initial tangential velocity:

We will use the initial velocity perpendicular to the shell calculated on Earth, but adding on the initial tangential velocity will give us a new resultant initial velocity.
Because the tangential and perpendicular velocities are at right angles to each other, we need to find the angle with which the resultant velocity is going.

The angle between direction of the motion and tangent to the surface is

Now that we have the angle, we look at the following diagram to find the max distance from the shell of the cylinder h'.

First we find r' and then subtract of r' from r to get h':

The ball reaches 9.908 meters high in the cylinder.