Chapter 7.4, Problem 41PS

Solution

a.

To calculate : The distance does the ball travel between the first and second bounce and between second and third bounce.

The ball travels $12$ feet between the first and second bounce, $9$ feet between the second and third bounce.

Given information :

A student drops a rubber ball from a height of $8$ feet and each time ball hit ground, it bounces $75\%$ of its previous height.

Formula used :

Calculate the sum of total distance covered by ball between the consecutive bounces.

Calculation :

It can be observed that ball fell from a height of $8$ feet and bounces back for the first time.

After that it travels up for $6$ feet and falls down $6$ feet and bounces again.

The distance between first and second bounce for the ball is $6+6=12$ feet.

From, the picture it can be seen that between second and third bounce the ball travels $4.5$ feet up and the $4.5$ feet down, then bounces back.

The distance between second and third bounce for the ball is $4.5+4.5=9$ feet.

Thus, the ball travels $12$ feet between the first and second bounce, $9$ feet between the second and third bounce.

b.

To calculate : The infinite series to model total distance travelled by the ball if the distance travelled before the first bounce is excluded.

The infinite series to model total distance travelled by the ball is $2\cdot {\displaystyle \sum _{i=1}^{\infty }6{\left(0.75\right)}^{i-1}}$ .

Given information :

A student drops a rubber ball from a height of $8$ feet and each time ball hit ground, it bounces $75\%$ of its previous height.

Formula used :

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

Calculation :

If the distance before first bounce is ignored, the ball travels distance of $2\left(6\right)$ in the second bounce.

So, total distance covered by the ball in the subsequent bounces can be written as follows.

  $\begin{array}{c}2\cdot 6+2\cdot 4.5+2\cdot 3.375+2\cdot 2.531+\cdots =2\cdot \left(6+4.5+3.375+2.531+\cdots \right)\\ =2\cdot \left(6+6\left(0.75\right)+6{\left(0.75\right)}^2+6{\left(0.75\right)}^3+\cdots \right)\end{array}$

Here the first term is $a_1$ is $6$ and $r$ is $0.75$ .

The geometric series is written as $a_n=2\cdot {\displaystyle \sum _{i=1}^{\infty }6{\left(0.75\right)}^{i-1}}$ .

Thus, the infinite series to model total distance travelled by the ball is $2\cdot {\displaystyle \sum _{i=1}^{\infty }6{\left(0.75\right)}^{i-1}}$ .

c.

To calculate : The distance travelled by the ball including the distance travelled before the first bounce.

The distance travelled by the ball including the distance travelled before the first bounce is $56$ feet.

Given information :

A student drops a rubber ball from a height of $8$ feet and each time ball hit ground, it bounces $75\%$ of its previous height.

Formula used :

The sum of infinite geometric series with first term $a_1$ and common ration $r$ is calculated using the formula $S=\frac{a_1}{1-r}$ .

Here, $\left|r\right|<1$ . If $\left|r\right|\ge 1$ , then the series has no sum.

Calculation :

The total distance travelled if the distance travelled before the first bounce included is $8+2\cdot {\displaystyle \sum _{i=1}^{\infty }6{\left(0.75\right)}^{i-1}}$ .

For the geometric series ${\displaystyle \sum _{i=1}^{\infty }6{\left(0.75\right)}^{i-1}}$ , the value of $a_1$ is $6$ and $r$ is $0.75$ .

So,

  $\begin{array}{c}8+2\cdot {\displaystyle \sum _{i=1}^{\infty }6{\left(0.75\right)}^{i-1}}=8+\frac{2\left(6\right)}{1-0.75}\\ =8+\frac{12}{0.25}\\ =8+48\\ =56\end{array}$

Thus, the distance travelled by the ball including the distance travelled before the first bounce is $56$ feet.

d.

To prove : The distance travelled by the ball before the first bounce is $7h$ feet if the ball is dropped from a height of $h$ feet.

Given information :

A student drops a rubber ball from a height of $8$ feet and each time ball hit ground, it bounces $75\%$ of its previous height.

Formula used :

The sum of infinite geometric series with first term $a_1$ and common ration $r$ is calculated using the formula $S=\frac{a_1}{1-r}$ .

Here, $\left|r\right|<1$ . If $\left|r\right|\ge 1$ , then the series has no sum.

Explanation :

If the distance covered is $h$ , then the total distance travelled if the distance travelled before the first bounce included is $h+2\cdot {\displaystyle \sum _{i=1}^{\infty }\left(h\cdot 0.75\right){\left(0.75\right)}^{i-1}}$ .

For the geometric series ${\displaystyle \sum _{i=1}^{\infty }\left(h\cdot 0.75\right){\left(0.75\right)}^{i-1}}$ , the value of $a_1$ is $0.75h$ and $r$ is $0.75$ .

So,

  $\begin{array}{c}h+2\cdot {\displaystyle \sum _{i=1}^{\infty }\left(h\cdot 0.75\right){\left(0.75\right)}^{i-1}}=h+\frac{2\left(0.75h\right)}{1-0.75}\\ =h+\frac{2\left(0.75h\right)}{0.25}\\ =h+2\left(3h\right)\\ =7h\end{array}$