Question

4. a) A ping pong ball has a 75% rebound ratio. When you drop it from a height of k feet, it bounces and bounces endlessly. If the height of the bounce follows geometric progression, find the general term that represent the situation in term of k. Hence, if the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces and the total distance travelled by the ball when it strikes the ground for 12th time ​

Answer 1

First part of question:

Find the general term that represents the situation in terms of k.

The general term for geometric series is:

$a_{n}=a_{1}r^{n-1}$

$a_{1}$ = the first term of the series

$r$ = the geometric ratio

$a_{1}$ would represent the height at which the ball is first dropped. Therefore:

$a_{1} = k$

We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:

$r=\frac{3}{4}$

Our general term would be:

$a_{n}=a_{1}r^{n-1}$

$a_{n}=k(\frac{3}{4}) ^{n-1}$

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

$k$ represents the initial height:

$k = 235\ ft$

$n$ represents the number of times the ball bounces:

$n = 6$

Plugging this back into our general term of the geometric series:

$a_{n}=k(\frac{3}{4}) ^{n-1}$

$a_{n}=235(\frac{3}{4}) ^{6-1}$

$a_{n}=235(\frac{3}{4}) ^{5}$

$a_{n}=55.8\ ft$

$a_{n}$ represents the highest height of the ball after 6 bounces.

Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. ​

This would be easier to solve if we have a general term for the sum of a geometric series, which is:

$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$

We already know these variables:

$a_{1}= k = 235\ ft$

$r=\frac{3}{4}$

$n = 12$

Therefore:

$S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }$

$S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }$

$S_{n}=(4)(235)(1-\frac{3}{4} ^{12})$

$S_{n}=910.22\ ft$