Question

A ball is released from a height of $x$ meters. Given that it bounces back to half of its original height. Find the product of heights after eight bounces. [4 MARKS]

Answer 1

Formula: 1 Mark
Application: 1 Mark
Answer: 2 Marks

Given that

The ball is released from a height of $xm$

Height reached after first bounce $=\frac{x}{2}$

Height reached after second bounce $=\frac{x}{4}=\frac{x}{2^2}$

Height reached after third bounce $=\frac{x}{8}=\frac{x}{2^3}$

Height reached after fourth bounce $=\frac{x}{16}=\frac{x}{2^4}$

Height reached after fifth bounce $=\frac{x}{32}=\frac{x}{2^5}$

Height reached after sixth bounce $=\frac{x}{64}=\frac{x}{2^6}$

Height reached after seventh bounce $=\frac{x}{128}=\frac{x}{2^7}$

Height reached after eight bounce $=\frac{x}{256}=\frac{x}{2^8}$

$\therefore$ Product of first 9 heights $=x\times \frac{x}{2}\times \frac{x}{2^2}\times \frac{x}{2^3}\times \frac{x}{2^4}\times \frac{x}{2^5}\times \frac{x}{2^6}\times \frac{x}{2^7}\times \frac{x}{2^8}$
$=\frac{x^{(1+1+1+1+1+1+1+1+1)}}{2^{(1+2+3+4+5+6+7+8)}}$
$(\because a^b\times a^c=a^{b+c})$
$=\frac{x^9}{2^{36}}$

The product of heights after 8 bounces $=\frac{x^9}{2^{36}}$