Answer
a) $8.125$ft
b) $x \approx 16.33 ft$
Work Step by Step
a) Given: $f(x)=-\dfrac{32}{(20)^2}x^2+x+5$
or, $f(x)=-\dfrac{-2}{25}x^2+x+5$
But $a=-\dfrac{-2}{25} \lt 0$ shows that the function has a maximum value.
The maximum value occurs at $x=-\dfrac{b}{2a}=-\dfrac{1}{2(\dfrac{-2}{25})}=6.25$
Thus, the maximum value of the function $f(x)$ at $x=6.25$ is:
$f(6.25)=-\dfrac{-2}{25}(6.25)^2+6.25+5$
This gives:
$f(6.25)=8.125$ ft
b) Consider $f(x)=0$
$-\dfrac{2}{25}x^2+x+5=0$
$x=\dfrac{-1\pm \sqrt {1+\dfrac{8}{5}}}{2(\dfrac{-2}{25})}$
This gives:
$x \approx 16.33 ft$ or, $x \approx -3.83 ft$
Since, $x \gt 0$ so we will only consider the positive solution.