In the situation you describe where the height of the ball is
reduced by half with each bounce, the energy of each bounce is
half the energy of the previous bounce, since the potential
energy is directly proportional to the height of the bounce, m
and g being constant. This means that the kinetic energy which is
given by 1/2*m*v2 is reduced by half. If we look at
the kinetic energy in one bounce compared to the kinetic energy
of the next bounce,
(1/2*m*v12)/(1/2*m*v22)=2
or,
v12/v22=2
or,
v12=2*v22
or,
v1=1.414*v2
rather than
v1=2*v2
as you suggest.
I would look at the problem this way:
A 1 meter height after the third bounce implies a 2 meter height
after the second bounce, a 4 meter height after the first bounce
and a 8 meter initial height. Use the relationship between
distance fallen and time of fall to calculate the time of the
initial 8 meter drop. Then, since a ball on the way up is subject
to exactly the same acceleration as a ball on its way down,
recognize that the rise time for each bounce will equal the fall
time. So calculate the fall time for a 4 meter fall and double
it, the fall time for a 2 meter fall and double that, and finally
the time for a 1 meter fall not doubled since we stop the problem
with the ball at the 1 meter height. Adding all these times
together should give you the answer.
Hope this helps.
J. D. Jones
M. Casco Associates