Answer
The total distance that the ball covers before it comes to rest is 180 feet.
Work Step by Step
The model can be represented by a geometric sequence with $a_{1} = 20$ and $r = \frac{4}{5}$ which the general term $a_{n} = 20(\frac{4}{5})^{n-1}$ where n is the number of dropping
Since the ball rebounds after the first dropping, the distance traveled will be double at each of the subsequent rebound after the first, hence, the total distance that the ball covers before it comes to rest is
= $(\frac{20}{1 - \frac{4}{5}}) \times 2$ - $20$
= $100 \times 2 - 20$
= $180$ feet
