Hint 4

The Coriolis force on a falling particle is

$${\bf F}_{\rm cor}=2\omega gt\cos\lambda\,{\bf e}_1,$$

as discussed in the lectures. Neglecting effects of order $\omega^2$, this can be integrated twice with respect to $t$ to get the deflection when the particle has fallen a distance $h$:

$$x_E \approx {g\over 3} \left[ {2h\over g} \right]^{3/2} \omega \, \cos\lambda,$$

in the e$_1$ direction (i.e. eastwards). This is in the same direction as the Earth's motion. Why?

For the numbers given, the deflection is about 1.5 cm.