Question

If a rubber ball consistently bounces back $\frac{2}{3}$ of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped?

• A. $\frac{16}{81}$
• B. $\frac{16}{27}$
• C. $\frac{4}{9}$
• D. $\frac{37}{81}$
• E. $\frac{9}{12}$

Answer 1

The correct option is A $\frac{16}{81}$

Each time the ball is dropped and it bounces back, it reaches $\frac{2}{3}$ of the height it was dropped from.

After the first bounce, the ball reach $\frac{2}{3}$ of the height from which it was dropped - let us call it the original height.

After the second bounce, the ball with reach $\frac{2}{3}$ of the height it would have reached after the first bounce.

So at the end of the second bounce, the ball would have reached $\frac{2}{3}\times \frac{2}{3}$ of the original height $=\frac{4}{9}$ the of the original height.

After the third bounce, the ball with reach $\frac{2}{3}$ of the height it would have reached after the second bounce.

So, at the end of the third bounce, the ball would have reached $\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}$ the of the original height.

After the fourth and last bounce, the ball with reach $\frac{2}{3}$ of the height it would have reached after the third bounce.

So, at the end of the last bounce, the ball would have reached $\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}$ of the original height $=\frac{16}{81}$ of the original height.