When a ball bounces, it rises to $\frac{3}{4}$ of the height from which it fell. If the ball is dropped from a height of 32 m, how high will it rise at the third bounce?

### Concept & Formula Each bounce reduces the height to a constant fraction of its previous height. This forms a Geometric Progression (GP) sequence. If the initial height is $H$ and the bounce retention fraction is $r$, the height after $n$ bounces is: $$\text{Height after } n \text{ bounces} = H \times r^n$$ ### Step-by-Step Solution * **Given:** Initial height ($H$) = 32 m Bounce fraction ($r$) = $\frac{3}{4}$ Number of bounces ($n$) = 3 * **Calculate the height after the 3rd bounce:** $$\text{Height} = 32 \times \left(\frac{3}{4}\right)^3$$ $$\text{Height} = 32 \times \frac{27}{64}$$ * **Simplify the expression:** $$\text{Height} = \frac{32}{64} \times 27 = \frac{1}{2} \times 27 = \frac{27}{2} = 13\frac{1}{2} \text{ m}$$ ### Exam Strategy & Shortcut Write down the chain multiplication directly: $32 \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}$. Cancel out factors step-by-step: * After 1st bounce: $32 \times \frac{3}{4} = 24$ * After 2nd bounce: $24 \times \frac{3}{4} = 18$ * After 3rd bounce: $18 \times \frac{3}{4} = \frac{54}{4} = 13.5 = 13\frac{1}{2}$ ### Common Pitfall Calculating the height *before* the third bounce (i.e., stopping at the second bounce calculation) or multiplying the reduction factor incorrectly by doing $32 \times \frac{3}{4} \times 3$. ### Final Answer **Therefore, the correct answer is 13\frac{1}{2} m.**