a ball is dropped from a height of 48m and rebounds two-third of the distance it falls. If it continues to fall and rebound in this way,how far will it travel before coming to rest?

Answer 1

Make a little diagram of what actually happens.
After the first bounce, the ball goes up and down 2/3 of it previous bounce, but the first event can be considered only half a bounce.

so I am going to consider this to be a geometric series, pretending the first bounce is complete. Just have to remember to subtract 48 from the final answer.

so sum = 96 + 96(2/3) + 96(2/3)^2 + ...
a = 96, r = 2/3

sum(all the terms) = a/(1-r)
= 96/(1- 2/3)
= 96/(1/3)
= 288
But this includes the first 48 m, which did not happen.

total distance = 288-48 = 240 m

or

total distance
= 48 + 2(48)(2/3) + 2(48)(2/3)^2 + ...
= 48 + 96(2/3)[ 1 + 2/3 + (2/3)^2 + ...]
= 48 + 64 [1/(1-2/3)]
= 48 + 64[1/3]
= 240