Perhaps this might help?
Hi,
Perhaps the following experiment might help.
Think about being on a merry-go-round. If you are close to the center and leaped straight up into the air, your horizontal speed would not be very great - you would land on the merry-go-round almost at the point from where you jumped. However, if you are near the edge of the merry-go-round and jumped upwards, your horizontal speed would be much greater since the edge of the merry-go-round is moving faster than a point near the center (via v_T=r\omega where v_T is the tangential velocity, r is the radius, and \omega is the angular velocity). When you come down from your leap, you most likely will not land on the merry-go-round because your faster horizontal velocity will have carried you off the edge. In other words, the farther you are from the axis of rotation, the faster the tangential velocity will be (at the same angular velocity of the merry-go-round, of course).
Now transfer this experiment to a tower on the Earth. At the Earth's surface, the tangential velocity has some numerical value given by v_T(R)=R\omega where R is now the length of the line perpendicular to the axis of rotation at your latitude value, say \theta. Then R=r_\text{Earth}\cos\theta where r_\text{Earth} is the radius of the Earth. If the tower has a height of h, then the top of the tower is farther from the axis of rotation than its base, now given by R_h=(r_\text{Earth}+h)\cos\theta. Note that R_h>R. Therefore the tangential velocity at the top of the tower is greater than at its base. To be specific, the equations give the tangential velocity at the top v_T(R_h)=R_h\omega>R\omega=v_T(R) greater than the tangential velocity at the base. So an object dropped from the top of a tower is traveling faster tangentially than the base of the tower. Hence, as it falls, its horizontal velocity (neglecting air resistance) must stay the same as it was at the tower top, and this horizontal velocity is greater than the tower base's horizontal velocity. Thus the object "outruns" the base --- that is, it falls towards the East!
A proper mathematical description of this effect will involve a term looking like -2\vec{\omega}\times\vec{v} where \times is not the normal multiplication between two real numbers but rather is the cross product between two vectors: the \vec{\omega} vector is the vector pointing in the direction of the axis of rotation (towards Polaris, in other words) and \vec{v} is the vector pointing in the direction of the velocity of the object, in this case in the vertical downward direction. The cross product of two vectors is perpendicular to both, and by the right-hand rule, would point in the Eastward direction in both the Northern and Southern Hemispheres. So, an object dropped from a tower falls East of the vertically straight down location. By the way, this is the same reason why if you shoot a rifle vertically straight up, the bullet will land to the West of your location -- believe it or not, this little experiment was performed at the Picatinny Arsenal in NJ in the 1930s, I believe. They shot a rifle vertically in the center of a small lake on a calm day to see where the bullet splashed down --- I believe they never could locate its landing position.
If you want help figuring out how the -2\vec{\omega}\times\vec{v} term (called Coriolis) arises, just ask.
Good luck,
Syd
P.S. I'm new at this...hope my answer wasn't too detailed or longwinded.
