Chapter 4.6, Problem 75PPSE

Solution

a.

To Find: The function in vertex form. Find the maximum height of the punt.

The maximum height is $36.81\text{ft}$

Given:

  $h=-0.01x^2+1.18x+2$

Kickers foot $=x$

Height of the ball $=h$

Explanation:

From the function, $h=-0.01x^2+1.18x+2$, $a>0$ is the minimum value of $y,$ and $a<0$ is the maximum value of $y$ .

It is given that $a=-0.01$ , so solve for the maximum value after that.

Find the $x$ using this formula, to calculate the maximum value or the maximum height,

  $x=\frac{-b}{2a}$

Substitute $b=1.18$ and $a=-0.01$

  $\begin{array}{l}x=\frac{-1.18}{\left(2\right)\times \left(-0.01\right)}\\ x=59\end{array}$

Substitute $x=59$ in the function $h=-0.01x^2+1.18x+2$:

  $\begin{array}{c}h=-0.01x^2+1.18+2\\ =-0.01{\left(59\right)}^2+1.18\left(59\right)+2\\ =-34.81+69.62+2\\ =36.81\end{array}$

Thus, the maximum height is $36.81\text{ft}$

b.

To Find: The height the player can reach to block the punt if the nearest defensive player is $5\text{ft}$ horizontally from the point of impact.

The player must reach $7.65\text{ft}$ to stop the punt.

Given:

  $h=-0.01x^2+1.18x+2$

Kickers foot $=x$

Height of the ball $=h$

Explanation:

To block the punt, the player must reach as high as the height of ball.

Given the ball's height:

  $h=-0.01x^2+1.18x+2$

Since the player is $5$ feet away from kicker's foot, so

  $x=5$

So, find the ball's height when $x=5$:

So, substitute $x=5$ in the function:

  $\begin{array}{c}h=-0.01{\left(5\right)}^2+1.18\left(5\right)+2\\ =-0.25+5.9+2\\ =7.65\end{array}$

So, to stop the punt, the player must reach $7.65\text{ft}$.

c.

To Find: The distance the ball would go down the field before it hit the ground, if suppose the ball was not blocked but continued on its path.

The distance the ball would go down the field before it hit the ground is $120\text{ft}$ .

Given:

  $h=-0.01x^2+1.18x+2$

Kickers foot $=x$

Height of the ball $=h$

Explanation:

From the given function $h=-0.01x^2+1.18x+2$ , find $x$ by using the quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=-0.01,b=1.18$ and $c=2$ :

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(1.18\right)\pm \sqrt{{1.18}^2-4\left(-0.01\right)\left(2\right)}}{2\left(-0.01\right)}\\ =\frac{-1.18\pm \sqrt{1.3924+0.08}}{-0.02}\\ =\frac{-1.18\pm \sqrt{1.4724}}{-0.02}\\ =-1.67,120\end{array}$

Eliminate the negative value, since the distance cannot be negative.

So, the distance the ball would go down the field before it hit the ground is $120\text{ft}$ .