How to calculate the horizontal shift?

[centinela20]

Homework Statement

I have a homework problem where we consider a point over the earth's surface, which is in a different parallel to the equator. There is a centrifugal force due to the rotation of the earth. (there is an attached figure of this).
I need to calculate the horizontal shift due to a_{cx} of an object falls vertically from 50 m above this point.

Relevant Equations

a_{cy} = \omega^2 R \cos{\theta}^{2} and
a_{cx} =\omega^2 R \sin{\theta} \cos{\theta}

I don't understand what horizontal shift means. I believe that means calculate the horizontal component of a_cx. But when I do that a_{cx} is in terms only of the angle and the radius of the earth. But what I need is to include the 50 m vertical distance, so I think that maybe we need to use a_{cy} too and calculate the net acceleration acting at this point. But I don't really know. I am pretty lost.

 

[haruspex]

what horizontal shift means

I would take it as meaning displacement, hence the need to specify the height.

Btw, you do not mean "There is a centripetal force due to the rotation of the earth." In the rotating frame there is a centrifugal force. Centripetal force is usually considered in an inertial frame, but it is not an applied force; it is the radial component of the net of the applied forces.

 

[centinela20]

I would take it as meaning displacement, hence the need to specify the height.

Btw, you do not mean "There is a centripetal force due to the rotation of the earth." In the rotating frame there is a centrifugal force. Centripetal force is usually considered in an inertial frame, but it is not an applied force; it is the radial component of the net of the applied forces.

Yes, sorry that was a typo mistake.
So if I take it as a displacement it would be $\triangle x = p_{final} - p_{initial}$
and the initial postition is the 50 m,
is the final position $x= \frac{1}{2} a_{cx}t^2$ ?

 

[haruspex]

Yes, sorry that was a typo mistake.
So if I take it as a displacement it would be $\triangle x = p_{final} - p_{initial}$
and the initial postition is the 50 m,
is the final position $x= \frac{1}{2} a_{cx}t^2$ ?

No, horizontal displacement, i.e. how far it lands from where it would have landed if not for Earth's rotation.