Recent content by centinela20

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...

Hello, I'm Centinela. I'm a student. I hope this forum help me and I'll help everyone any time I can.