Answer
a) $\frac{65}{8}$ ft.
b) $16.33$ ft.
Work Step by Step
(a) Let's rewrite our function down:
$y=-\frac{32}{20^2}x^2+x+5$
As we can see, this function follows the form of a quadratic function $f(x)=ax^2+bx+c$, where $a=-\frac{32}{20^2}$, $b=1$ and $c=5$. Therefore, we'll use the properties of a quadratic function.
Because $a<0$, this function's maximum value occurs at $x=\frac{-b}{2a}$. And since both variables $a$ and $b$ are known, we can calculate $x$ and therefore obtain our maximum value.
$x=\frac{-b}{2a}=\frac{-1}{2*(-\frac{32}{20^2})}=\frac{1}{\frac{2*32}{400}}=\frac{400}{2*32}=\frac{200}{32}=\frac{25}{4}$
$f(\frac{25}{4})=-\frac{32}{20^2}(\frac{25}{4})^2+\frac{25}{4}+5=\frac{-32*25^2}{4^2*20^2}+\frac{25}{4}+\frac{20}{4}=\frac{-2*25}{16}+\frac{45}{4}=\frac{-25}{8}+\frac{90}{8}=\frac{65}{8}$
Therefore, the answer to a) is $65/8$ ft.
(b) asks us to solve for the horizontal distance ($x$) when the ball hits the ground. To calculate $x$, we need to find $y$ in this scenario, which is the height of the ball from the ground; and because the ball is touching the ground, $y$ is now equal to $0$. Thus:
$\frac{-32}{400}x^2+x+5=0$
$\frac{-2}{25}x^2+x+5=0$
Using the Quadratic Formula:
$x=(-b+\sqrt {b^2-4ac})/2a$ or $x=(-b-\sqrt {b^2-4ac})/2a$
$x=\frac{-1+\sqrt {1-4*\frac{-2}{25}*5}}{\frac{-4}{25}}$ or $x=\frac{-1-\sqrt {1-4*\frac{-2}{25}*5}}{\frac{-4}{25}}$
$x= \frac{25-25\sqrt {2.6}}{4}$ or $\frac{25+25\sqrt {2.6}}{4}$
$x=-3.83$ or $x=16.33$
Because distance traveled can't be negative, our answer for b) is $16.33$ ft.
