# Thread: Another sequence and serie problem !

1. ## Another sequence and serie problem !

The height of each bounce of a rubber ball is 2/3 of the height of the previous bounce. The ball is dropped from a height of 15m. What distance does it travel ( as it moves vertically up and down ) from the moment that it touches the ground for the first time till it touches the ground for the seventh time?

Struggeling here again!

Thanks people , appreciate it !

FuNk

2. Originally Posted by FuNkY14
The height of each bounce of a rubber ball is 2/3 of the height of the previous bounce. The ball is dropped from a height of 15m. What distance does it travel ( as it moves vertically up and down ) from the moment that it touches the ground for the first time till it touches the ground for the seventh time?...
Hello,

1. The heights which were reached by the ball perfom a geometric sequence:
$\displaystyle h(n)=10 \cdot \left( \frac{2}{3} \right)^n$

2. Every bounce adds two equal distances (one way up, same way down) to the total distance.

3. When the ball touches the ground the 7th time, it has performed 6 complete bounces.

4. The total distance is a geometric serie:
$\displaystyle s(n)=\frac{a(q^n -1)}{q-1}$

Plug in the values you know:
$\displaystyle s(n)=\frac{10 \cdot \left( \left(\frac{2}{3} \right)^6 -1 \right)}{\left( \frac{2}{3} -1\right)}$
You'll get: $\displaystyle s(6)=\frac{6650}{243}$

You have to double this value (see 2.) and you'll get approximately 54.73

Greetings

EB

3. Hey thx man!
I assume that the 10 there is actually the 15m?

Thx!

4. Originally Posted by FuNkY14
Hey thx man!
I assume that the 10 there is actually the 15m?
Thx!
Hello,

when the ball falls down from 15 m, it'll reach after the first touch down (2/3)*15m=10m. So I ignored the 15 m, because the first complete bounce is going from zero to 10 and back to zero. Can you see it now?

According to your problem, the initial 15 m are not part of the sequence or serie.

Greetings

EB

5. O yes. I get you now!
Thank you!!