Total Distance Travelled by a Ball Dropped Comes to Rest

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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