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## ?
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...