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 total distance travelled by the ball from the time it was dropped until it eventually comes to rest on the floor.

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

Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Divisibility Proof Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Proof Pythagoras Theorem Quadratic Quadratic Factorise Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume