At the equator a stationary (relative to the Earth) body falls down from the height \(h=500 \mathrm{~m}\). 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.
1.117. Will deviate to the east by the distance \(x \approx\) \(\approx \frac{2}{3} \omega h \sqrt{2 h / g}=24 \mathrm{~cm}\). Here \(\omega\) is the angular velocity of the Earth's rotation about its axis.