Question

This image shows that when a football is kicked, the nearest defensive player is 6 feet from the kicker's foot. The height of the punted football, f(x), in feet, can be modeled by the equation. Round all answers 2 decimals....... a. What is the maximum height of the punt?b. How far from the kicker's foot does the maximum height occur?c. How far must the nearest defensive player, who is 6 feet from the kicker's foot, reach to block the punt; in other words, what is the value of f(6)?d. If the ball is not blocked by the defensive player, how far down the field will it go before hitting the ground?

Answer 1

Given:

$f(x)=-0.015x^2+1.25x+2$

For the maximum :

$\begin{gathered} f^{\prime}(x)=0 \\ f(x)=-0.015x^2+1.25x+2 \\ f^{\prime}(x)=-0.015\times2x+1.25 \\ f^{\prime}(x)=-0.03x+1.25 \end{gathered}$

For the maxima point :

$\begin{gathered} -0.03x+1.25=0 \\ x=\frac{1.25}{0.03} \\ x=\frac{125}{3} \end{gathered}$

At x=125/3 football achived maximum height.

so maximu height is:

$\begin{gathered} f(x)=-0.015x^2+1.25x+2 \\ =-0.015\times(\frac{125}{3})^2+(1.25\times\frac{125}{3})+2 \\ =-26.04+52.08+2 \\ =28.04 \end{gathered}$

Maximum height achived is 28.04