F-LE Basketball Rebounds

According to Wikipedia, the International Basketball Federation (FIBA) requires that a basketball rebound to a height of 1300 mm when dropped from a height of 1800 mm.

Suppose you drop a basketball and the ratio of each rebound height to the previous rebound height is 1300:1800. Let $h$ be the function that assigns to $n$ the rebound height of the ball (in mm) on the $n^{\rm th}$ bounce or rebound. Complete the chart below, rounding to the nearest mm.

$n$ $h(n)$

0 1800

1

2

3

Write an expression for $h(n)$.

Solve an equation to determine on which bounce the basketball will first have a rebound height of less than 100 mm.

Commentary

This task involves a fairly straightforward decaying exponential. Filling out the table and developing the general formula is complicated only by the need to work with a fraction that requires decisions about rounding and precision.

Teachers can extend this task by having students engage in a related classic classroom activity. Students bounce a basketball beneath a sonar detector (such as the Calculator Based Ranger from Texas Instruments). The CBR and graphing calculator will generate a graph of the ball's position at times throughout the bouncing. Students can trace along this graph to find at least five rebound heights and then compute the average of the $\frac{h(n)}{h(n-1)}$ ratios. Students can then use this average along with their knowledge of the initial height of the ball to develop an exponential model for the data.

Solutions

Solution: Solution 1

On the first rebound, the ball must rise to a height of 1300 mm according to the IBF regulations.

To determine the second rebound height, 1300 mm must be multiplied by the required fraction $\frac{1300}{1800}=\frac{13}{18}$:

$$ 1300\left(\frac{13}{18}\right) = \frac{16,900}{18} \approx 939 \text{ mm} $$

For each succeeding rebound the previous rebound height is multiplied by $\frac{13}{18}$;

$n$ $h(n)$

0 1800

1 1300

2 $1300\left(\frac{13}{18}\right) \approx 939$

3 $938.9\left(\frac{13}{18}\right) \approx 678$

$n$	$h(n)$
0	1800
1	$1300=1800\left(\frac{1300}{1800}\right)$
2	$938.9 \approx 1300\left(\frac{13}{18}\right) =1300\left(\frac{13}{18}\right) \left(\frac{13}{18}\right) = 1800 \left(\frac{13}{18}\right)^2$
3	$678.1 \approx 938.9\left(\frac{13}{18}\right) \approx 1800 \left(\frac{13}{18}\right)^2 \cdot \left(\frac{13}{18}\right) = 1800 \left(\frac{13}{18}\right)^3 $
n	$1800\left(\frac{13}{18}\right)^n$

Generalizing from the table, we have $h(n) = 1800\left(\frac{13}{18}\right)^{n}$: this makes sense because by the $n$th rebound, we've multiplied by $\frac{13}{18}$, $n$ times.

The rebound on which the height will be approximately 100 mm can be determined by solving for $n$ given $h(n) = 100$: $$ 100 = 1800\left(\frac{13}{18}\right)^{n} $$ or $$ \frac{1}{18} = \left(\frac{13}{18}\right)^{n} $$ Logarithms may be used to solve such an equation: $$ \begin{align} \ln\left({\frac{1}{18}}\right) &= \ln\left({\frac{13}{18}}\right)^{n}\\ \ln\left({\frac{1}{18}}\right) &= n \cdot \ln\left(\frac{13}{18}\right) \end{align} $$ rounding yields the equation $$ -2.890 \approx n \cdot (-0.325) $$ which tells us $$ \begin{align} n &\approx 8.892 \ \end{align} $$ So, the first time the rebound will not be at least 100 mm with be on the 9th rebound.

Basketball Rebounds by illustrativemathematics, used under CC-BY-NC-SA 3.0

\(n\)	\(h(n)\)
0	1800
1	1300
2	\(1300\left(\frac{13}{18}\right) \approx 939\)
3	\(938.9\left(\frac{13}{18}\right) \approx 678\)

\(n\)	\(h(n)\)
0	1800
1	\(1300=1800\left(\frac{1300}{1800}\right)\)
2	\(938.9 \approx 1300\left(\frac{13}{18}\right) =1300\left(\frac{13}{18}\right) \left(\frac{13}{18}\right) = 1800 \left(\frac{13}{18}\right)^2\)
3	\(678.1 \approx 938.9\left(\frac{13}{18}\right) \approx 1800 \left(\frac{13}{18}\right)^2 \cdot \left(\frac{13}{18}\right) = 1800 \left(\frac{13}{18}\right)^3 \)
n	\(1800\left(\frac{13}{18}\right)^n\)