Exercise 2/99 from 6th Ed of Meriam, Kraige - Engineering reads:
An object which is released from rest from the top $A$ of a tower of height $h$ will appear not to fall straight down due to the effect of the earth's rotation. It may be shown that the object has an eastward horizontal acceleration relative to the horizontal surface of the earth equal to $2v_y \omega cos \gamma$, where $v_y$ is the free-fall downward velocity, $\omega$ is the angular velocity of the earth, and $\gamma$ is the latitude, north or south. Determine the deflection $b$ if $h$ = 1000 ft and $\gamma$ = 30° north.
The exercise itself is simple. What I want to know is how to show this eastward horizontal acceleration. How does one prove it?
I tried to check it. Using spherical coordinates:
Acceleration is
$$ \vec{a} = a_r \hat{r} + a_\phi \hat{\phi} + a_\theta \hat{\theta} $$
wherein its components are:
$$ a_r = \ddot{r} - r\dot{\theta}^2 - r\dot{\phi}^2 \cos^2 \theta $$ $$ a_\phi = r\ddot{\phi}\cos\theta + 2\dot{r}\dot{\phi}\cos\theta - 2r\dot{\phi}\dot{\theta}\sin\theta $$ $$a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} + r\dot{\phi}^2 \sin\theta \cos\theta$$
Now, if the object is falling relative to earth, its velocity is $\dot{r} = -v$. Earth rotation is $\dot{\phi}=\omega$ and it is constant ($\ddot{phi}=0$). Also, for a given latitude $\theta=\gamma$, $\dot{\theta}=\ddot{\theta}=0$. Also, gravitational acceleration is towards the center of the Earth, so $\ddot{r}=-g$. Substituting it in the components, we get:
$$ \vec{a} = \left(-g-r\omega^2 \cos^2 \gamma \right) \hat{r} + \left(-2 v \omega \cos \gamma\right) \hat{\phi} + \left( r \omega^2 \sin \gamma \cos \gamma \right) \hat{\theta} $$
Tangential component is $-2 v \omega \cos \gamma$, which is similar to the expected answer. However, the negative sign indicates this component is opposite to the direction of rotation of the earth, which is wrong. What I am doing wrong? How does one prove this eastward horizontal acceleration?
