Modeling with Exponential Functions

5.11.3 Modeling with Exponential Functions

5.11.3 • Modeling with Exponential Functions

Activity

Your teacher will give your group three different kinds of balls.

Your goal is to measure the rebound heights, model the relationship between the number of bounces and the heights, and compare the bounciness of the balls.

A close-up of various sports balls, including footballs, soccer balls, and a basketball, piled together. The central football has Pro Touch Touchdown printed on it.

1. Create a table similar to the one below and begin your investigation. Make sure to note which ball goes with which column.

$n$ , Number of Bounces	$a$ , Height for Ball 1 (cm)	$b$ , Height for Ball 2 (cm)	$c$ , Height for Ball 3 (cm)
0
1
2
3
4

Compare your answers: Sample Data:

$n$ , Number of Bounces	$a$ , Height for Ball 1 (cm)	$b$ , Height for Ball 2 (cm)	$c$ , Height for Ball 3 (cm)
0	150	150	150
1	80	100	60
2	41	70	25
3	20	50	13
4	11	35	6

2. Which one appears to be the bounciest? Which one appears to be the least bouncy? Be prepared to show your reasoning.

Compare your answer:

The second ball is the bounciest. The balls were all dropped from the same height, and its bounces are the highest. The third ball is the least bouncy.

3. For each one, write an equation expressing the bounce height in terms of the bounce number $n$ .

Compare your answers:

Ball 1: $a = 150 \cdot (\frac{1}{2})^{n}$ or $a = 150 \cdot (0.5)^{n}$
Ball 2: $b = 150 \cdot (\frac{7}{10})^{n}$ or $b = 150 \cdot (0.7)^{n}$
Ball 3: $c = 150 \cdot (\frac{2}{5})^{n}$ or $c = 150 (0.4)^{n}$

4. Explain how the equations could tell us which one is the most bouncy.

Compare your answers:

The bounciness of the ball is measured by the rebound factor, which is the base of the exponential expression. That is $\frac{1}{2}$ for the first ball, $\frac{7}{10}$ for the second ball, and $\frac{2}{5}$ for the third ball. The larger this number, the bouncier the ball.

5. If the bounciest one were dropped from a height of 300 cm, what equation would model its bounce height $h$ ?

Compare your answer:

$h = 300 \cdot (\frac{7}{10})^{n}$

6. Graph, on the same coordinate plane, the equation for each ball that models the bounce height $h$ after $n$ bounces using the Desmos tool below.

Compare your answer:

Three exponential decay curves are plotted on a graph, each starting at a high value and approaching zero as they move right. The curves are black, orange, and teal, positioned left to right, on a grid background.

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7. Use the graph of the equation that models the bounce height $h$ after $n$ bounces for all 3 balls to predict the difference in the height of Ball 1 after 5 bounces.

Compare your answer:

The difference in height after 5 bounces is 5.5. Using the equation: $h = 150 \cdot (0.50)^{n}$ and graphing $h = 150 \cdot (0.50)^{5}$ The graph shows the height to be 4.6875 after 5 bounces $(n = 5)$ .

This is consistent with the reasoning:

The height of the ball is decreasing by a factor of about $\frac{1}{2}$ after each bounce.
After the $4^{th}$ bounce, the height difference is 11. It should decrease by $\frac{1}{2}$ for each bounce after that. So, after 1 more bounce (the $5^{th}$ bounce), it will decrease by a factor of about $\frac{1}{2}$ . The $5^{th}$ bounce would be $11 \cdot \frac{1}{2} = 5.5$ .
4.6875 rounds to 5, and 5.5 rounds to 6.
Take the average of the two numbers. $\frac{5 + 6}{2} = 5.5$

8. Use the graph of the equation that models the bounce height $h$ after $n$ bounces for all 3 balls to predict the difference in the height of Ball 2 after 5 bounces.

Compare your answer:

The difference in height after 5 bounces is 25. Using the equation: $h = 150 \cdot (0.7)^{n}$ and graphing $h = 150 \cdot (0.7)^{5}$ . The graph shows the height to be 25.2105 after 5 bounces $(n = 5)$ .

This is consistent with the reasoning:

The height of the ball is decreasing by a factor of about $\frac{7}{10}$ after each bounce.
After the $4^{th}$ bounce, the height difference is 35. It should decrease by $\frac{7}{10}$ for each bounce after that. So, after 1 more bounce (the $5^{th}$ bounce), it will decrease by a factor of about $\frac{7}{10}$ . The $5^{th}$ bounce would be $35 \cdot \frac{7}{10} \approx 24.5$ .
Rounding either 25.2105 or 24.5 results in the answer questions 25

9. Use the graph of the equation that models the bounce height $h$ after $n$ bounces for all 3 balls to predict the difference in the height of Ball 3 after 5 bounces.

Compare your answer:

The difference in height after 5 bounces is 2. Using the equation: $h = 150 \cdot (0.4)^{n}$ and graphing $h = 150 \cdot (0.4)^{5}$ . The graph shows the height to be 1.636 after 5 bounces $(n = 5)$ .

This is consistent with the reasoning:

The height of the ball is decreasing by a factor of about $\frac{2}{5}$ after each bounce.
After the $4^{th}$ bounce, the height difference is 6. It should decrease by $\frac{2}{5}$ for each bounce after that. So, after 1 more bounce (the $5^{th}$ bounce), it will decrease by a factor of about $\frac{2}{5}$ . The $5^{th}$ bounce would be $6 \cdot \frac{2}{5} \approx 2.4$ .
Rounding either 1.636 or 2.4 results in the answer questions 2.

Are you ready for more?

Extending Your Thinking

1. If Ball 1 were dropped from a point that is twice as high, would its bounciness be greater, less, or the same? Be prepared to show your reasoning.

2. Ball 4 is half as bouncy as the least bouncy ball. What equation would describe its height $h$ in terms of the number of bounces $n$ ?

3. Ball 5 was dropped from a height of 150 centimeters. It bounced up very slightly once or twice and then began rolling. How would you describe its rebound factor? Be prepared to show your reasoning.

Compare your answer:

The same. The bounciness of the ball does not depend on the height from which it is dropped. This is why the successive rebound heights are modeled by an exponential function.
$h = 150 (\frac{1}{5})^{n}$
The rebound factor is very close to 0. Because the ball barely left the ground after hitting it, we could say that it lost almost all of its height with each bounce.