Equations of motion for non-rigid bodies I've been thinking about how we usually find the time required for a rigid body (like a baton) that's rotating to be thrown into the air and return to where it was thrown by treating it's CM as a point mass then using the standard equations of motion.
I was curious about throwing a body into the air which isn't rigid. If we have an object comprised of a finite number of connected masses but not rigid so the distance between them can change and it was thrown into the air, would this mean that the CM would change position within the object (I feel like it would) and how would you then use the equations of motion to find the time it would take to fall down?
Maybe it's a silly question but I feel like it wouldn't be as simple as using say $y = v_{0}t + \frac{1}{2}at^2$ (assuming $y_{0} = 0$) if the CM itself is fluid within the object as it moves? I hope this is understandable.
 A: The center of mass will follow the same trajectory regardless of the internal forces and motions of an object (as long as the external forces applied to the object don't change as the object changes).  Even if an object explodes mid-trajectory, the CM will still follow the original trajectory (although in practice for explosions, the air resistance will change, which is a change in the external force, so the trajectory will change).
Generally, then, to solve these problems one would calculate the CM trajectory, and then the internal motions separately.  It could be complex to estimate the exact touchdown time of a free fall, since the internal motions could make the object stretch or shrink and tweek the contact time, but the CM will contact at the same time.
(In a counter-factual where this wasn't true, maybe you could stand on a board and hold ropes attached to the board, then jump, in mid-flight, pull the board up to your feet, and jump again, and continue to do this to gain flight! But, sadly, what will happen in the real world is that the CM of you and the board will follow the trajectory of your initial jump, no matter what antics you do during that trajectory.)
