Answer
a) The maximum height attained by the shot is 33.66 feet, and the distance at which it occurs is 26.25 feet.
b) The maximum horizontal distance is $\underline{\text{55}\text{.3 feet}}$
c) The shot was released from $\underline{\text{6}\text{.1 feet}}$.
Work Step by Step
(a)
Compare the equation$f\left( x \right)=-0.04{{x}^{2}}+2.1x+6.1$ to the standard equation and get:
$a=-0.04,\text{ }b=2.1,\ \text{and }c=6.1$.
Since, $a<0,$ the x value of where the maximum height will occur is at $x=\frac{-b}{2a}$.
$\begin{align}
& x=\frac{-2.1}{2\left( -0.04 \right)} \\
& =\frac{-2.1}{-0.08} \\
& =26.25
\end{align}$
So, the distance where the maximum height of the shot occurs at 26.25 feet from the point of release.
The maximum height of the shot is:
$\begin{align}
& f\left( 26.25 \right)=-0.04{{\left( 26.25 \right)}^{2}}+2.1\left( 26.25 \right)+6.1 \\
& =33.70
\end{align}$
Therefore, the maximum height of the shot occurs at $\text{26}\text{.25 feet}$ from the point of release and the maximum height is $\text{33}\text{.65 feet}$.
(b)
The maximum horizontal displacement is determined by calculating the x-intercept.
To find the $x$ -intercept, replace $f\left( x \right)\text{ by 0}$ and solve the quadratic equation obtained.
$0=-0.04{{x}^{2}}+2.1x+6.1$
And solving the equation using the quadratic formula, we get,
$\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-2.1\pm \sqrt{{{\left( 2.1 \right)}^{2}}-4\left( -0.04 \right)\left( 6.1 \right)}}{2\left( -0.04 \right)} \\
& =\frac{-2.1\pm \sqrt{5.386}}{-0.08} \\
& =\frac{-2.1\pm 2.3}{-0.08}
\end{align}$
Now, the value thus obtained is $x=55.26\text{ or }x=-2.76$
As $x\ge 0,$ the horizontal distance of the shot will be $x=55.26\text{ feet}$.
Hence, the maximum horizontal distance is $55.3\text{ feet}$.
(c)
The height from which the shot is released is given by its y-intercept, that is $f\left( 0 \right)$.
So, $f\left( 0 \right)=-0.04{{\left( 0 \right)}^{2}}+2.1\left( 0 \right)+6.1$feet
Thus, the height of the shot will be $6.1\text{ feet}$.
Hence, the height at which the shot was released is $\text{6}\text{.1 feet}$.
