Question

A football is kicked toward an end zone with an initial vertical velocity of 30 ft/s. The function h(t) = -16+ 30t models the height h (in feet) of the football at time t (in seconds). Which statement about the height of the football is true? A.The football does not reach a height of 15ft B. The football reaches a height of exactly 15ft C.The football reaches a height that is greater than 15ft

Answer 1

Answer:

A. The football does not reach a height of 15ft

Step-by-step explanation:

Given

$h(t) = -16t^2 + 30t$

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:

$f(x) = ax^2 + bx + c$:

$f(\frac{-b}{2a}) = maximum/minimum$

i.e we first solve for $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for x in $f(x) = ax^2 + bx + c$

In our case:

First we need to solve $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for t in $h(t) = -16t^2 + 30t$

By comparison:

$b = 30$

$c = -16$

$\frac{-b}{2a} = \frac{-30}{2 * -16}$

$\frac{-b}{2a} = \frac{-30}{-32}$

$\frac{-b}{2a} = \frac{30}{32}$

$\frac{-b}{2a} = \frac{15}{16}$

Substitute $\frac{15}{16}$ for t in $h(t) = -16t^2 + 30t$

$h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})$

$h(\frac{15}{16}) = -16(\frac{225}{256}) + \frac{450}{16}$

$h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}$

$h(\frac{15}{16}) = \frac{-225 + 450}{16}$

$h(\frac{15}{16}) = \frac{225}{16}$

$h(\frac{15}{16}) = 14.0625$

This implies that the maximum height reached is 14.0625ft.

So, the option that answers the question is A because $14.0625 < 15$