Conservation Of Angular Momentum of a falling stone I am currently stuck on the following question. 
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 and in what direction? Solve this problem by conserving angular momentum. 
My current interpretation is that when the stone falls in order to converse angular momentum the angular velocity must increase. This leads to the stone travelling to the east. 
I equated the angular momentum at the start       L0=*(re+h)mVo
where re is radius of earth, h is the height above that, m is mass and Vo is inital velocity. To the angular momentum at any time
L=(re+h-1/2gt^2)m(Ve+Vo)
where the 1/2at^2 term comes from gravitational acceleration and Ve is the additional velocity. 
I rearranged to find Ve and then integrated over the time it takes to fall the 500m in order to work out the extra distance the stone traveled. However, my answer is 1/2 what the answer should be. I have checked the maths and it seems to be fine. So I believe my set up is wrong. Any help would be appreciated. 
 A: From conservation of angular momentum we can derive angular velocity as a function of height:
$$\omega(h) = \omega_0 \left(\frac{R + h_0}{R+h}\right)^2$$
where $\omega_o$ is the angular velocity of the earth, $R$ is the radius of the earth, $h$ is the current height and $h_0$ is the initial height.
The horizontal velocity (in the frame of reference of the earth) is 
$$v(h) = (\omega_0-\omega(h))(R+h) =\\
\omega_0(R+h) - \omega_0\frac{(R+h_0)^2}{(R+h)}$$
A bit more manipulation gives
$$v(h) = \frac{\omega_0}{R+h}\left((R+h)^2 - (R+h_0)^2 \right)\\
\approx\omega_0\frac{2R(h-h_0)}{R+h}\\
\approx2\omega_0(h - h_0)$$ 
(The approximations are valid because $R>>h$).
In other words, the velocity scales directly with the difference in height. As a function of time, we write
$$v(t) = 2\omega_0((h_0-\frac12gt^2)-h_0)\\
=-omega_0gt^2$$
Now we integrate:
$$x = \int_0^T - \omega_0 gt^2 dt\\
= -\frac13\omega_0gT^3$$
where $T$ is the time taken for the fall, roughly $T=\sqrt{\frac{2h_0}{g}}$
Substituting, we get
$$x = -\frac13 \omega_0g\left(\frac{2h_0}{g}\right)^\frac32=24 cm$$
Since you did not tell us what the "correct" answer was, or what the details of your math were, I can't tell whether this solves your problem...
