There are simulations programs around to solve this...

[Mickey]

It was interesting to know the answer and the explanation.

100 ants (zero-length points) walk on a meter stick (a line) at 1 cm/
second. When two ants collide, they both reverse direction. If an ant
reaches the end of the stick, it falls off. What arrangement of ants
maximizes the time before all ants have fallen
off? How long can they last?

Regards,
Jyoti

[Akhil Bhiwal]

Place each ant at 1 cm distance. 1st ant at 0cm, 2nd ant at 1cm and so
on. 50th ant at 49cm. No ant at 50cm. 51st ant at 51cm and so on till
100cm. Make the left 50 and right 50 ants facing the center(i.e.,
center).
Now, if we start the ant movements, the ant at 0 and 100cm fall first,
then ants at 1 and 99cm, and so on.
Max. time taken in this case will be by the ant 50 and 51. They will
last for 101 second.

Reason:
Let's assume there is a 4cm stick and 4 ants. In this case, first set
of ants fall at 3sec and second set of ants fall at 4sec.
Similarly, in case of 6cm stick and 6 ants, we get time as: 4, 5, 6.
Paralleling this with given question, I get the time for 1st set of
ants as 51, for 2nd set as 52, and so on...


Regards,
Akhil Bhiwal

[Mickey]

Akhil,

I think the correct answer is 100 seconds and the arrangement can be
something else also. I will wait till tomorrow to post the solution;
in case some one else wants to improve the solution.

Regards,
Jyoti

[Akhil Bhiwal]

@Jyoti: Yes, the correct answer is 100 seconds. I did a calculation
mistake previously.

Waiting for other solutions. :)

[Mickey]

Try getting the answer for one ant... and then two ants... and then
three ants... You would have got the answer for 100 ants by now.
The maximum time is 100s for any number ants. Be it 1 or 100 or 1000.

Here is the key:
When two ants collide it is the equivalent of them passing through
each other. With this insight it is indeed the case that the maximum
is 100 seconds for any number of ants.

Regards,
Jyoti