|  | 
			
			
			
					
					
			
				
					
						
	I mean... it is linear in speed,  it's just nonlinear in energy.
		
			
			
				
					  Originally Posted by OngBonga   I thought it was b/h, but that's because I assumed the scale from 0 to 1 was linear. 
 Max height of an object in projectile  motion is a measurement of it's total mechanical energy (simplified to Potential + Kinetic, here).  It is at rest  in a potential energy field at some distance h from what we've chosen  to call 0.  At height 0, we model that 100% of its potential energy is turned into kinetic energy.  That part of the model is linear in energy.
 
 At the elastic collision, we have a speed that is before and after.  The COR is linear in this speed ratio, with 1 being a perfectly elastic collision, and 0 being a perfectly inelastic collision.
 
 Then we're back into the projectile motion regime and it's linear in energy again, with total energy constant.
 
 
 
	
		
			
			
				
					  Originally Posted by OngBonga   But still, assuming ideal conditions with perfect elasticity and no other influences, the answer is still a COR of 1, I believe. The ball bounces to precisely to the same location from where it was dropped. Fun fact - the ground also falls slightly towards the ball, and then bounces back the same distance it fell, which gives it a COR of 1 too.  Rereading the original question, you got it.
 Saying the ball is perfectly elastic is one thing, but the ground itself wasn't said, and the ball-ground interaction is what's under scrutiny.
 But if we assume the ground is perfectly rigid, then the ball being perfectly elastic seals the deal.  COR = 1.
 
 It's kinda a trick question because perfect elasticity means COR = 1, and knowing how to calculate COR when b != h is a red herring.
 
 
 
 
	By assuming the ground's mass is "much much greater" than the ball, we also assume it's much much greater than our own mass, as well.
		
			
			
				
					  Originally Posted by OngBonga   I don't believe so. We have three masses, because "you" are dropping the ball. In effect, we're assuming the ground's mass is infinite; it is un-accelerate-able, and a suitable place to attach a Newtonian inertial reference frame.
 
 Which, as you say, is an absurd approximation.
 Or is it?
 The Earth's mass is dozens of orders of magnitude higher than my mass, which is assumed to be within a couple orders of magnitude of the ball.
 
 If we're assuming the reason the ball accelerates is due to gravity, then the assumption that we're on a roughly Earth-sized planet seems apt.  Under those conditions, the fault in our approximation will show up only dozens of decimal places deep.  So we're technically wrong, but not measurably wrong.  So the wrongness is quantified as "acceptable."
 
 
 
	In that sense, all knowledge is absurd.
		
			
			
				
					  Originally Posted by OngBonga   Even if we have perfect elasticity, an absurd concept in its own right, we have a third gravity influence altering the velocity of both the ball and ground. 
 Perfect elasticity isn't entirely absurd, no more than any other model (guess / approximation) of reality.
 
 Think of 2 magnets' like poles repelling each other.
 Mostly, what is deformed that is the origin of the force?
 The magnetic field.
 The (electro)magnetic field is perfectly elastic.  Any energy put into deforming it is given back in full.
 
 
 
	Precisely 0 is perfectly inelastic, when the ball sticks to the wall.
		
			
			
				
					  Originally Posted by OngBonga   Infinite mass? The COR of a ball "bouncing" off an infinite mass is surely precisely zero. The "after" speed is 0, and that goes in the numerator of the COR fraction.
 
 
 
	Provided the given statement that the balls a perfectly elastic, yes.
		
			
			
				
					  Originally Posted by OngBonga   See above. It does matter because other objects in the universe mean non-uniform gravitational influences. But if we assume a perfectly isolated system, with a ghost dropping the ball onto a much much larger ball (not actually important, balls can be same size), made of the same perfectly elastic material with perfectly uniform density, and everything else being perfect so there is only ball ground and bounce, then it's COR=1 surely. 
 If the origin of the force that causes the acceleration during the bounce is the deformation of macroscopic objects (like balls), then the COR will always be less than 1.  The deformation will cause some internal heating, which is energy lost from the mechanical interaction.
 |