First of all, the rotational speed of the surface of the surface of the earth is more like v = 465 meters per second, not 3000 kilometers per second. The fundamental principle here is the conservation of angular momentum. At the surface of the earth the angular momentum of a body of mass m is L = mvR where R is the radius of the earth. As you ascend to a height, h, your tangential velocity decreases by a factor of δv in order to have Rv = (R+h)*(v- δv). The earth seems to move a bit faster than you and you would land a very small distance to the west. An integral is involved in order to get the exact answer.
Newton predicted in 1679 that if you dropped a ball from a height of 8 meters it would land about a half millimeter to the east (the algebraic sign of h changes).
I think perhaps you are asking a little different question than the one Lee answered. Let's look at the situation in a non-rotating reference frame at rest with respect to the center of the earth. Initially you and the surface of the earth are moving in the same direction (east) at the same speed. When you jump, you pick up some upward velocity, but that does nothing to reduce your sideways velocity
. So you continue to east. The direction in which gravity pulls you does
change, because as you move east the direction from you to the center of the earth changes. Then Lee's calculations take care of the small effects that arise from your changing distance from the earth.
(published on 11/10/11)