Question

A ball is dropped from a height of 3 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.
(a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter your answer as an improper fraction.)

Answer 1

When the ball hits the ground the first time it will have bounced up a third of the initial 3 feet, which is 1 foot. Similarly, the second time it bounces, it will have bounced a third of 1, which is 1/3. The third time, it will bounce 1/9.

This question is asking for the total distance the ball traveled so:
Drop 1 = 3 ft.
Bounce after drop 1 = .3 ft
Drop 2 = .3 ft
Bounce after drop 2 = .1 ft
Drop 3 = .1 ft
Bounce after drop 3 = .03 ft
Drop 4 = .03 ft.

The instant the ball hits the ground for the 4th time is after drop 4. Adding these distances up, we get:
3+ .3+.3+.1+.1+.03+.03 = approximately 3.9 ft

Answer 2

Correction:
I skipped from 3 ft to 1/3 ft (skipping 1 ft) by mistake

Keeping these as fractions, your answer should actually be
\[3+1+1+1/3+1/3+1/9+1/9=53/9\]