Answer:
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Formula: 1 Mark
Application: 1 Mark
Answer: 2Marks
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$=\frac{x^9}{2^{36}}$