A 33300 lb (earth weight) rocket in outer space has two constant forces acting upon it,

Question

F1 = (-166i + 392j) lb and
F2 = (50i + -54j) lb.
Its initial velocity is (39i + 64j) ft/s.
How far is it from where it started after 33 seconds?

Answer 1

The net force is F1 + F2 = -116 i + 338 j

In the x direction, the final location is changed by 39*33 + (1/2)(-116/33,300)*33^2

There is a similar equation for the change in the y coordinate: Voy*t + (1/2)(Fy/m)t^2

Answer 2

Thanks for the timely help.