Question

A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is dropped from a height of 12 feet, how far will it have traveled when it hits the surface the fifth time?

Answer 1

Answer:

$\frac{259}{54}\text{ or }4.8\text{feet}$

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one $\frac{1}{6}$ as high as the preceding one.

TO FIND: If this ball is dropped from a height of $12$ feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces $\frac{1}{6}$ of preceding one and comes down the same distance.

When the ball is dropped from $12$ feet height

after first hit $=12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4$

new height $=12\times\frac{1}{6}=2\text{feet}$

after second hit $=2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}$

new height $=2\times\frac{1}{6}=\frac{1}{3}\text{ feet}$

after third hit  $=\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}$

new height $=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}$

after fourth hit $=\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}$

adding all distance $=4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}$

                                 $=\frac{259}{54}$ feet

Hence the ball will travel   $\frac{259}{54}$ feet before it hits the surface fifth time.