Word/probability problem please help! :)

[1a2s3d4f5g6h7j8k9l]

hi all!

Here's the problem:
If a ball bounces .8 the distance it is dropped, how many bounces will the ball make before it rises to less than one foot if the ball is dropped from a height of 6 feet?
I've tried to solve it...trying to divide stuff by 6 or 0.8...clearly I don't understand the problem! help...

Thanks!

 

[Unknown008]

Okay, it goes like this:

Drop from 6;
Bounce to 6(0.8) = 4.8
Bounce to 4.8(0.8) = 3.84
Bounce to 3.84(0.8) = 3.072
etc

This is a geometric progression, where this formula applies:

\(\displaystyle T_n = ar^{n-1}[/math]

a is the first term, 4.8, Tn is the bounced height, r is the common ratio 0.8, after n bounces.

So, after 1 bounce, you have \(\displaystyle T_1 = 4.8(0.8)^{1-1} = 4.8\)

After 2 bounces \(\displaystyle T_2 = 4.8(0.8)^{2-1} = 3.84\)

So, now, solve:

\(\displaystyle 4.8(0.8)^{n-1} < 1\)\)

 

[soroban]

Hello, 1a2s3d4f5g6h7j8k9l!

"Divide by 6 or 0.8" . . . What are you smoking?

If a ball bounces 0.8 the distance it is dropped, and is dropped from a height of 6 feet,
how many bounces will the ball make before it rises less than one foot?

This is a geometric sequence.

The first term is \(\displaystyle a_1 = 6\), the common ratio is \(\displaystyle r = 0.8\)
. . The \(\displaystyle n^{th}\) term is: .\(\displaystyle a_n \:=\:a_1r^{n-1}\)

Our general term is: .\(\displaystyle a_n \:=\:6(0.8)^{n-1}\)
. . which we want to be less than 1.

We have: .\(\displaystyle 6(0.8)^{n-1} \;<\;1 \quad\Rightarrow\quad 0.8^{n-1} \;<\;\frac{1}{6}
\)

Take logs: .\(\displaystyle \ln\left(0.8^{n-1}\right) \;<\;\ln(6^{-1}) \quad\Rightarrow\quad (n-1)\ln0.8 \;<\;-\ln6 \)

Divide by \(\displaystyle \ln0.8\), which is negative: .\(\displaystyle n-1 \;>\;-\dfrac{\ln6}{\ln0.8}\)

And we have: .\(\displaystyle n \;>\;\dfrac{\ln6}{\ln0.8} + 1 \;=\;9.029626932\)


Therefore, it will make \(\displaystyle n = 10\) bounces.

 

[Unknown008]

Hello Soroban,

I believe that the first term is 4.8 and not 6, because of the fact that:
After 1 bounce, the height the ball reaches is 4.8, right? (or in other terms, the ball will have made 1 bounce, before it reaches 4.8 m as its maximum point)

Substituting that into the formula using 6 as first term and n = 1 would give:

6(0.8)^{1-1} = 6(0.8)^0 = 6(1) = 6 m

Obviously, the ball does not bounce at a height of 6 m after one bounce...

And furthermore, doing the problem using brute force yields:

After n bounce:OOOOOOHeight Reached:
OOOO1OOOOOOOOOOOt4.8
OOOO2OOOOOOOOOOOt3.84
OOOO3OOOOOOOOOOOt3.072
OOOO4OOOOOOOOOOOt2.4576
OOOO5OOOOOOOOOOOt1.96608
OOOO6OOOOOOOOOOOt1.572864
OOOO7OOOOOOOOOOOt1.2582912
OOOO8OOOOOOOOOOOt1.00663296
OOOO9OOOOOOOOOOOt0.805306368