Question

The height a ball bounces is less than the height of the previous bounce due to friction. The heights of the bounces form a geometric sequence. Suppose a ball is dropped from one meter and rebounds to 95% of the height of the previous bounce. What is the total distance traveled by the ball when it comes to rest?

-Does the problem give you enough info to solve the problem?
-How can you write the general term of the sequence?
-What formula should you use to calculate the total distance?

Answer 1

you drop the ball and it goes down one meter
lets put that 1 aside for a moment
then it goes up \(.95\) and therefore down \(.95\) for a total distance of \(2\times .95^2\)

Answer 2

ok that was wrong
it goes up \(.95\) and down \(.95\) for a total of \(2\times .95\)

Answer 3

then it goes up \(.95^2\) and down \(.95^2\) for a total of \(2\times .95^2\)

Answer 4

leaving the 1 out of it, you get
\[2\times .95+2\times .95^2+2\times .95^3+2\times .95^4+...\]

Answer 5

that is the geometric series you need to add up, then add the 1 at the end

Answer 6

i am still kind of confused. so thats the series. where do i put the 1? and what formula do i use for the total distance?

Answer 7

@satellite73

Answer 8

leave the one aside, add it at the end
it doesn't fit the pattern

Answer 9

the sum is found by adding a geometric series
\[2(.95+.95^2+.95^3+...)=2\times \frac{.95}{1-.95}\]

Answer 10

i will try it...

Answer 11

it is actually easier to do it without a calculator
\[\frac{.95}{1-.95}=\frac{.95}{.05}=\frac{95}{5}=19\]

Answer 12

multiply it by 2 and get 38, then add the 1 and get 39

Answer 13

this still makes no sense. that isnt even the formula the book uses to find the sum. its \[S _{n}=\frac{a _{1}(1-r ^{n})}{ 1-r }\] where a1 is the first term and r is the common ratio and n is the number of terms.

Answer 14

@timo86m can you help?

Answer 15

use geometic sequence

Answer 16

1+r^1+r^2+....+r^n
r = .95