The football coach threw a football from a platform to his
quarterback below. The height of the football, h, at time t seconds is modeled by the equation h(t) = -16t^2 + 28t + 15.
1. What is the maximum height of the ball?
2. If the quarterback caught the ball at a height of 6 feet, how many seconds was the ball in the air?
3. Give the domain and range of the function.

Answer:

1)27.25 feet

2)-0.2774,2.027

3)Domain = All real numbers

Range = y∈R : [tex]y \leq \frac{109}{4}[/tex]

Step-by-step explanation:

The height of the football, h, at time t seconds is modeled by the equation [tex]h(t) = -16t^2 + 28t + 15[/tex]

General quadratic equation : [tex]ax^2=bx+c=0[/tex]

1) Maximum height will be at [tex]\frac{-b}{2a}=\frac{-28}{2(-16)}=\frac{7}{8}[/tex]

To find maximum height Substitute [tex]t = \frac{7}{8}[/tex] in the given equation

[tex]h(t)=-16(\frac{7}{8})^2+28(\frac{7}{8})+15\\h(t)=27.25 feet[/tex]

2)

Substitute h(t)=6

So, [tex]-16t^2+28t+15=6[/tex]

[tex]-16t^2+28t+9=0\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\t=\frac{-28\pm \sqrt{28^2-4(-16)(9)}}{2(-16)}\\t=\frac{-28+ \sqrt{28^2-4(-16)(9)}}{2(-16)},\frac{-28- \sqrt{28^2-4(-16)(9)}}{2(-16)}\\t=-0.2774,2.027[/tex]

3)

Domain = All real numbers

Range = y∈R : [tex]y \leq \frac{109}{4}[/tex]

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