Question 8
After $n$ bounces, the rebound height (cm) of a ball, $t_n$, is modelled by the rule $t_n=240\times0.5^{(n-1)}$. Calculate the difference in rebound height (cm) between the first bounce and the third bounce.
Not quite. One way to get 90 is to use $0.5^n$ instead of $0.5^{(n-1)}$. That makes the first bounce 120 and the third 30, and $120-30=90$. With $n-1$ in the power, the first bounce is the full 240 cm. In 2024, 12.5% of Queensland students chose (A).
Not quite. 120 is the difference between the first and second bounces, $240-120$, and it is also the height of the second bounce. The question compares the first bounce with the third. In 2024, 15.6% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2024, 65.9% of Queensland students chose (C).
Not quite. One way to get 210 is $240-30$, the first bounce minus the fourth. The third bounce is $n=3$, so its height is $240\times0.5^2=60$. In 2024, 5.3% of Queensland students chose (D).
The bars are the rebound heights from the rule, drawn to scale. Choose a bounce to compare with the first. The orange gap is the difference.
Each bounce reaches half the height of the one before. Pick the bounce the question wants to compare with the first.
This is the question: first bounce minus third bounce. The gap is 180 cm, option (C).
First and second bounces. The gap is 120 cm, option (B), and the second bounce is 120 cm high as well, so 120 tempts you twice. The question asked for the third bounce.
First and fourth bounces gives 210 cm, option (D). That is one bounce too far.
This is one way to get option (A). With the power $n$ every bar is half the height it should be, so the first bounce comes out as 120 cm. The rule’s $n-1$ is there to make the first bounce exactly 240 cm.
With the power $n$ instead of $n-1$, every bar is half the height it should be. Check any rule by substituting $n=1$ first: this one must give 240 cm.
The first bounce is $n=1$. Substitute it into the rule:
$$t_1=240\times0.5^{(1-1)}=240\times0.5^0=240$$
Anything to the power 0 is 1, so the first bounce is 240 cm. That is what the $n-1$ is for.
The third bounce is $n=3$:
$$t_3=240\times0.5^{(3-1)}=240\times0.25=60$$
Check by halving: 240, then 120, then 60. Three bounces, three numbers.
Difference means subtract, larger minus smaller:
$$t_1-t_3=240-60=180\text{ cm}$$
When each height halves, the gap between two bounces in a row equals the lower height, so 120 is both. If you land on 120, ask whether you have found the gap between the first and third bounces or only the first and second.
$$\begin{aligned}t_1&=240\times0.5^{0}=240\\t_3&=240\times0.5^{2}=60\\t_1-t_3&=240-60=180\text{ cm}\end{aligned}$$
Substitute $n=1$ and $n=3$ into the rule as it is written, then subtract.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2024 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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