Challenging summation prob: A ball falls 10 ft and then boun

[auroralbutterfly]

This is the problem that was given to me:

A ball is dropped from a height of 10 feet. Each time it bounces back up, it bounces 0.65 times as high as the previous bounce. What is the total distance traveled by the ball? Also, write the summation notation.

I know that this is a geometric sequence with a common ratio of 0.65. I know that this should be an infinite geometric series also. This is what I've tried:

The ball falls 10 ft and then bounces back up 6.5 ft. So that means it bouncing down and then back up once is 16.5 ft.

The ball continues to bounce and bounces back down 6.5 ft and then goes back up 4.2 ft, which mean it'll be 16.5 + 6.5 + 4.2 = 27.2 ft

It goes down 4.2 ft and goes up 2.75 ft, so: 27.2 + 4.2 + 2.75 = 34.15 ft

It can go on and on until the ball stops. I know the answer is 47.14 ft so I can check my answer. But how can I find the answer easier and also write the summation notation?

Thank you so much for you time!

 

[arthur ohlsten]

return ht = 10+10[.65]+10[.65]^2+....10[.65]^n
distance= 10+2*10[a+a^2+a^3+...a^n] where a=.65
add and subtract 10
distance = 20+20[a+a^2+...a^n] -10
distance=20[1+a+a^2+..a^n] - 10

Sn=1+a+a^2+...a^n multiply by a
aSn=a+a^2+a^3+...a^(n+1) subtract aSn from Sn
Sn[1-a]=1-a^(n+1)
Sn=[1-a^(n+1)] / [1-a] for a=.65 and n-->oo
Sn=1/.35

distance=[20/.35] - 10
distance= [20-3.5]/.35
distance=1650/35
distance=47.1 ft. answer

Arthur

 

[soroban]

Re: Challenging summation prob: A ball falls 10 ft and then

Hello, auroralbutterfly!

A ball is dropped from a height of 10 feet.
Each time it bounces back up, it bounces 0.65 times as high as the previous bounce.
What is the total distance traveled by the ball?
Also, write the summation notation.

You already worked out all the necessary factrs . . . now put them together.

It falls \(\displaystyle 10\) feet.
It bounces up \(\displaystyle 10(0.65)\) feet, and falls \(\displaystyle 10(0.65)\) feet.
It bounces up \(\displaystyle 10(0.65)^2\) feet, and falls \(\displaystyle 10(0.65)^2\) feet.
It bounces up \(\displaystyle 10(0.65)^3\) feet, and falls \(\displaystyle 10(0.65)^3\) feet . . . and so on.

The total distance is:
. . \(\displaystyle \L\:S \:=\:10\,+\,\underbrace{2\cdot10(0.65)\,+\,2\cdot10(0.65)^2\,+\,2\cdot10(0.65)^3\,+\,\cdots}_{\text{geometric series}}\)

The geometric series has first term \(\displaystyle a\,=\,20(0.65)\) and common ratio \(\displaystyle r\,=\,0.65\)

Its sum is: \(\displaystyle \L\:K\:=\:\frac{20(0.65)}{1\,-\,0.65}\:=\:37.142857\)


Therefore: \(\displaystyle \L\:S\;=\;10\,+\,K\;\approx\;47.14\) feet.

 

[auroralbutterfly]

thank you so much to both of you!!

i understood it more~~
i'll be having a quiz tomorrow with a similar question, so hopefuly, i'll do well on it =)
thanks again!

 

[arthur ohlsten]

you are welcome.
The only thing a little different about your problem was that it onlu traveled 10ft once, and all other distansec twice, the rebound and subsequent fall.
Arthur