Question

Suppose you drop a ball from a window 45 meters above the ground. The ball bounces to 65% of its previous height with each bounce. What is the total number of meters (rounded to the nearest tenth) the ball travels between the time it is dropped and the 10th bounce?

Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio.

Answer 1

Sn= a1(1-rn)/1-r

Answer 2

You want to find the 10th bounce, which is the 10th term of the sequence.
The zero term is 45.
The first term is 45(0.65)
So the ratio is (0.65)

Using your formula
\[S _{n} = a _{1} \frac{ r^{n}-1 }{ r-1 }\]

You can calculate the first term, put in the ratio for r, and put in the 10th term for n.

Answer 3

yea this is a geometric series with common ratio r = 0.65

Answer 4

we have to remember that the distance travelled at each bounce = twice the distance so the first term is 2* 45* 0.65 = 58.5
so a1 = 58.5

so total distance after 10 bounces is 58.5 * (1 - 0.65^10)
------------ + 45
(1 - 0.65)

Answer 5

as a matter of interest we can work out the total distance the ball travels until its stops
using the simple formula S = a1 / (1 - r) because as n approaches infinity 0.65^n approaches zero

so this tptal will be 45 + 58.5 / (1 - 0.65) = 45 + 167.14 = 212.14 m

this geometric series converges.