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

Here `v_y=0` so we can take `y=0`, thus we get for the motion in the x-y plane.
`underset(..)x=-2omegav_zcostheta`
and `overset(..)z=-g`
Integrating, `z=-1/2g t^2`
`overset(.)x=omegagcosvarphit^2`
So `x=1/3omegagcosvarphit^3=1/3omegagcosvarphi((2h)/(g))^(3//2)`
`=(2omegah)/(3)cosvarphisqrt((2h)/(g))`
There is thus a displacement to the east of
`2/3xx(2pi)/(8)64xx500xx1xxsqrt((400)/(9*8))~~26cm`.