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.

At the equator a stationary (relative to the Earth) body falls down fr

1.	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» 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`.

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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.

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