A ball is dropped from a height of \( 16 \) metres onto a timber floor and bounces. After each bounce, the maximum height reached by the ball is \( 80 \% \) of the previous maximum height. Thus, after it first hits the floor, it reaches a height of \( 12.8 \) metres before falling again, and after the second bounce, it reaches a height of \( 10.24 \) metres before falling again.

## Part 1

Find the maximum height reached after the third bounce.

\( 10.24 \times 0.8 = 8.192 \text{ metres}\)

## Part 2

What kind of sequence is formed by the successive maximum height?

\( \text{Geometric Sequence with the first term of 20 metres and the common ratio of 0.8} \)

## Part 3

Find the ball’s total distance from when it was dropped until it eventually comes to rest on the floor.

\( \require{AMSsymbols} \begin{align} \text{initial height dropped} &= 16 \\ \text{first pair of equal up and down} & = 2 \times 16 \times 0.8 \\ \text{second pair of equal up and down} & = 2 \times 16 \times 0.8^2 \\ \text{third pair of equal up and down} & = 2 \times 16 \times 0.8^3 \\ &\cdots \\ \text{total distance} &= 16 + 2 \times 16 \times (0.8 + 0.8^2 + 0.8^3 + \cdots) \\ &= 16 + 2 \times 16 \times \displaystyle \color{red}{\frac{0.8}{1-0.8}} &\color{red}{S_{\infty} = \frac{\text{first term}}{1-\text{common ratio}} } \\ &= 144 \text{ metres} \end{align} \)

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