Re: A question concerning a ball?

I first tried to solve this problem by determining the various energies 
(i.e. potential, kinetic, and elastic) the ball would have during its 
bouncing.  I assumed that the ball would behave similar to a spring, with 
a specific spring constant 'k', using the equation:  E = 1/2kx^2.  Upon 
further thinking I realized this method overlooked that the "spring-like" 
factor of the ball is acting "ideally", or without any losses.  Another 
variable would have to be introduced to account for losses.  This, however,
would make the problem unnecessarily complex.

I discussed this problem with a friend of mine [Kent Nickerson, B.Sc. 
(Physics; U. of Waterloo, Canada), M.Eng. (Elec.Eng.; McMaster U., Canada)]
and he had a novel solution, which I will explain below:

First assume the following:
Ball radius = r
Initial drop height = h(0) (with respect to centre of ball)
Assume h(0) >>r, so that kinetic energy when ball hits ground is nearly
mgh(0)

Introduce an elastic coefficient, 'B', that can have a value between 0 and
1 (0 having no "bounciness" and 1 being perfectly bouncy).  'B' is the 
part of kinetic energy that is preserved after a bounce.  This can be done 
since no data is given to derive this.

Working this out then....after the first bounce, the ball will only reach 
'B' times the ball's original height, or:

		h(1) = B*h(0)

The same will happen for each consecutive bounce:

		h(2) = B*h(1)
		h(3) = B*h(2)
		h(4) = B*h(3)
		....
		h(n) = B*h(n-1)

Putting all of these little equatons together we get (after 'n' bounces):

	h(n) = B*B*B*B*B*....(there are 'n' B's)....*B*B*B*B*B*h(0)

Since there were 'n' terms, there are also 'n' B's.  Thus all of the 'B' 
terms in the previous equation can be writen as B^n.
Rewriting:
		h(n) = h(0)*B^n

Now, B^n is equivalent to exp(n*lnB) (you can check this in a calculus 
text), so a final solution can be written as:

	       -----------------------	
	      | h(n) = h(0)*exp(n*lnB) |
	       -----------------------


This answer is fairly simplified, but I think it is sufficient.  I hope 
this helps you.

[Just as an aside, I want to mention that this problem was one of the first
that Newton tackled with his "new" differential calculus.  The elastic coefficient, 
'B', that I refer to, Newton called the 'coefficient of restitution'.  There
should be some more information on this in some mechanics or calculus texts.]


	Kurt Frost
	kfrost@sympatico.ca