Answer:
A. The football does not reach a height of 15ft
Step-by-step explanation:
Given
[tex]h(t) = -16t^2 + 30t[/tex]
Required
Determine which of the options is true
The option illustrates the height reached by the ball.
To solve this, we make use of maximum of a function
For a function f(x)
Such that:
[tex]f(x) = ax^2 + bx + c[/tex]:
[tex]f(\frac{-b}{2a}) = maximum/minimum[/tex]
i.e we first solve for [tex]\frac{-b}{2a}[/tex]
Then substitute [tex]\frac{-b}{2a}[/tex] for x in [tex]f(x) = ax^2 + bx + c[/tex]
In our case:
First we need to solve [tex]\frac{-b}{2a}[/tex]
Then substitute [tex]\frac{-b}{2a}[/tex] for t in [tex]h(t) = -16t^2 + 30t[/tex]
By comparison:
[tex]b = 30[/tex]
[tex]c = -16[/tex]
[tex]\frac{-b}{2a} = \frac{-30}{2 * -16}[/tex]
[tex]\frac{-b}{2a} = \frac{-30}{-32}[/tex]
[tex]\frac{-b}{2a} = \frac{30}{32}[/tex]
[tex]\frac{-b}{2a} = \frac{15}{16}[/tex]
Substitute [tex]\frac{15}{16}[/tex] for t in [tex]h(t) = -16t^2 + 30t[/tex]
[tex]h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})[/tex]
[tex]h(\frac{15}{16}) = -16(\frac{225}{256}) + \frac{450}{16}[/tex]
[tex]h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}[/tex]
[tex]h(\frac{15}{16}) = \frac{-225 + 450}{16}[/tex]
[tex]h(\frac{15}{16}) = \frac{225}{16}[/tex]
[tex]h(\frac{15}{16}) = 14.0625[/tex]
This implies that the maximum height reached is 14.0625ft.
So, the option that answers the question is A because [tex]14.0625 < 15[/tex]
