Question

At the equator a stationary (relative to the Earth) body falls down from the height $h=500m$ Assuming the air drag to be negligible, find how much off the vertical, and in what direction, the body will deviate when it hits the ground.

Answer 1

$\ddot{x}=2m\omega (v_y\sin \phi -v_z\cos \phi )$, $\ddot{y}=0$ and $\ddot{z}=-g$
Here $v_y=0$ so we can take $y=0$, thus we get for the motion in $x\dot{z}$ plane.
$\ddot{x}=-2\omega v_z\cos \theta$
and $\ddot{z}=-g$
Integrating, $z=-\frac{1}{2}g{t}^2$
$\dot{x}=\omega g\cos \phi t^2$
So $x=\frac{1}{3}\omega g\cos \phi t^3=\frac{1}{3}\omega g\cos \phi {\left(\frac{2h}{g}\right)}^{\frac{3}{2}}$
$=\frac{2\omega h}{3}\cos \phi \sqrt{\frac{2h}{g}}$
There is thus a displacement to the east of
$\frac{2}{3}\times \frac{2\pi }{8}64\times 500\times 1\times \sqrt{\frac{400}{9.8}}\approx 26cm$