Even supposing a constant downwards gravitation acceleration if you do the math of your problem you get a four degree polynomial (parabola-circumference intersection): Suppose you are on a latitude with the radius $r$ to the rotation axis, and $\omega$ earth angular speed. You throw the corp vertically with speed $v$. We consider an inertial system with center on earth axis and same latitude, y axis in the vertical direction the corp is thrown and x toward the tangent speed $r\omega$. The corp motion law in this system is $y(t)=-\frac 12gt^2+vt+r$, $ x(t)=\omega rt$ so if you want to know where it lands you have to impose that it is on the circumference: $(-\frac 12gt^2+vt+r)^2+ \omega^2r^2=r^2$, which is a 4 grade equation. Suppose $\lambda_0$ to be the root we are interested in, then you can find the landing position relative to the starting one in radiant $\Delta\theta=\tan^{-1}\frac{y(\lambda_0)}{x(\lambda_0)}-\omega\lambda_0$. If you consider very long distance you should even consider the change in the direction of the gravitational force, so even more complex (supposing this can be done without escaping earth gravitational field).
