Dirac1238 said:
yes but what if the 2D square was rotating, would the equation of rotation be different then for let's say a 1D line.
GOOD QUESTION!
As long as we disregard air resistance, then a thrown object will conserve angular momentum, because the only acting force upon the object, gravity, works at the C.M of the object.
Thus, whatever energy associated with the object's rotation initially will be the same during the whole object's flight.
We can, therefore, ignore the object's rotational state when calculating its trajectory.
However, and this is important:
Air resistance is IMMENSELY important in order to describe the actual orbit of, say, a rotating baseball.
This is because the rotation of the ball creates a velocity differential in the ambient air, and therefore, a pressure differential upon itself as well.
This means that in a viscous fluid like air, a rotating ball will get quite a different course than the one predicted for a ball in vaccuum, rotating or not.