Solution:
Let MN = 18 m be the banner shaft and its shadow be LM = 9.6 m.
The distance of the highest point of the post, N from the far end, L of the shadow is LN.
In right calculated ∆LMN,
\[LN2\text{ }=\text{ }LM2\text{ }+\text{ }MN2\text{ }[by\text{ }Pythagoras\text{ }theorem]\]
\[\Rightarrow LN2\text{ }=\text{ }\left( 9.6 \right)2\text{ }+\text{ }\left( 18 \right)2\]
\[\Rightarrow LN2\text{ }=\text{ }9.216\text{ }+\text{ }324\]
\[\Rightarrow LN2\text{ }=\text{ }416.16\]
\[\therefore LN\text{ }=\text{ }\surd 416.16\text{ }=\text{ }20.4\text{ }m\]
Thus, the necessary distance is 20.4 m