After being dropped a ball always bounces back to \(\frac{2}{9}\)of the height of its previous bounce. After the first bounce it reaches a height of 260 inches. How high will it reach on bounce number 5?

0.63 inches


This problem is a series of multiplications by \(\frac{2}{9} \):
\(260 \text { inches } \times \frac{2}{9} \times \frac{2}{9} \times \frac{2}{9} \times \frac{2}{9}=\frac{260 \times 2 \times 2 \times 2 \times 2}{9 \times 9 \times 9 \times 9}=\frac{4160}{6561}=0.63 \text { inches }\)

Visit our website for other ASVAB topics now!