A problem on determining coriolis force on frames First of all I am defining x axis in east of object and y to be in the north of object . Let us suppose a body is falling from a height; then after falling a certain height indeed gain some velocity in the direction which I am defining in the local height of the observer in z direction.  indeed in a rest frame outside Earth it would not have velocity in y or x direction  but in a frame standing still on earth (perhaps moving with earth) it will gain a velocity in the east direction and y direction so when I determine the coriolis force for a certain moment then to determine the coriolis force while doing the matrix of 2$m\omega \times v$should I put $\vec v$ is equal to $\dot x\hat{i} + \dot y \hat{j} + \dot z \hat{k}$ as in the in the earth rotating frame it has velocity in the x and y directions but in the rest frame outside earth it never has velocity in x or y direction. So should I put  $\vec v$ =$0\hat{i} +0\hat{j} + \dot z\hat{k}$ ?  Show my exact question is that what should I take as the velocity to detect the velocity as the perception as in perception of a rest frame outside Earth or should I take the velocity of a observer who is in the earth frame who can't perceive his own rotation so sees the ball shifting. 
 A: The equivalent of the Newton's second law in the noninertial frame is expressed as
$$m\ddot{\vec r} = \vec{F}' - 2m \vec{\omega} \times\vec{v}.$$
Here everything is written in terms of the rotating coordinate system. $F'$ is the actual net force on the system, but it is written to be a function of the rotating coordinates. It also absorbed the centrifugal force, so in your case $\vec{F}' = m\vec{g}$ but $|\vec{g}| \neq 9.8$ N/kg.
So, the point here is that $\vec{v}$ is taken to be in the rotating coordinates -- is the viewpoint of an observer fixed on the Earth. 
Thus for the initial velocity you have $\vec{v}_0 = \dot{z}\hat{k}$. As the object begins to fall, it will become deflected from the Coriolis effect, and this velocity will in accordance with $\vec{v} \times \vec{\omega}$. However for the Earth, $\omega$ is quite small at $\sim 10^{-4}$ s$^{-1}$. Thus, to a first approximation:
$$m\ddot{\vec r} = \vec{F}' - 2m \vec{\omega} \times\vec{v}_0,$$
which will allow you to calculate the longitudinal deflection (in the x-direction).
As given in Goldstein p. 174, if you have that text, the corresponding inertial frame is a coordinate system which is stationary with respect to "local stars".
