I think the following substitutions allow it to be simply solved as a
one line solution
for some i between 1 and n
t(i) = sqrt ( d(i) / (0.5 * g)) - lets call this 1
where t(i) is the time for the ith ball to fall
d(i) is the height of the ith ball.
t(0) + t(1) .. + t(n) = T - lets call this 2
where T is the total time which is supplied.
Substituting for t(i) from 1 into 2
(sqrt(d(1)) + sqrt(d(2)) .... + sqrt(d(n)))/ sqrt(0.5 * g) = T
g = (sqrt(d(1)) + sqrt(d(2)) .... + sqrt(d(n)))**2 / (T**2 * 0.5)
where x**n = x raised to n.
hence it collapses down to a single variable.
On Sun, Aug 1, 2010 at 7:28 PM, Michael Mellinger
<[address removed]> wrote:
> I added a Perl solution. ?It seems I missed something in the problem
> because my code is much longer. ?No ugly Perl shortcuts taken:
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