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

`(259)/(54)\text{ or }4.8\text{feet}`

Explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one
`(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
`(1)/(6)` of preceding one and comes down the same distance.

When the ball is dropped from
`12` feet height

after first hit
`=12*(1)/(6)\text{ up}+12*(1)/(6)\text{ down}=4`

new height
`=12*(1)/(6)=2\text{feet}`

after second hit
`=2*(1)/(6)\text{ up}+2*(1)/(6)\text{ down}=(2)/(3)`

new height
`=2*(1)/(6)=(1)/(3)\text{ feet}`

after third hit
`=(1)/(3)*(1)/(6)\text{ up}+(1)/(3)*(1)/(6)\text{ down}=(1)/(9)`

new height
`=(1)/(3)*(1)/(6)=(1)/(18)\text{ feet}`

after fourth hit
`=(1)/(18)*(1)/(6)\text{ up}+(1)/(18)*(1)/(6)\text{ down}=(1)/(54)`

adding all distance
`=4+(2)/(3)+(1)/(9)+(1)/(54)`

`=(259)/(54)` feet

Hence the ball will travel
`(259)/(54)` feet before it hits the surface fifth time.