# Conservation of angular momentum of a falling particle

## Homework Statement

A stone is dropped from a stationary helicopter 500m above the ground, at the equator. How far from the point vertically below the helicopter does it land?

## Homework Equations

Conversation of AM

## The Attempt at a Solution

Let the height above the ground it is dropped be h, the radius of the Earth R, the mass of the stone m (which will cancel) and the angular velocity of the earth w. Then the angular momentum as it's dropped is mw(R+h)^2.

When the particle is at a height y above the Earth's surface, the stone has angular momentum m(R+y)(v_x+v_0) where v_0=(R+h)w is the velocity (in x direction) when it dropped, due to the helicopter being stationary, w.r.t. the Earth.

Now, y=h-0.5g*t^2, and Conservation of AM implies

mw(R+h)^2=m(R+h-0.5g*t^2)(v_x+(R+h)w).

I rearranged for v_x and integrated between t=0 and t'=Sqrt(2h/g), the time for the stone to hit the ground.

I go the answer x=12cm, but it should be x=24cm. Am i performing the integration incorrectly or have I set up the equations wrong?

Thanks, Rupe