Question

If a football is kicked straight up with an initial velocity of 96 ft/sec from a height of 4ft, then its height above the earth is a function of time given by h left parenthesis t right parenthesis equals negative 16 t squared plus 96 t plus 4h(t)=−16t^2+96t+4.

Answer 1

The height of the football above the ground as a function of time \( h(t) \) is given by the equation

\[
h(t) = -16t^2 + 96t + 4
\]

where:
- \( h(t) \) is the height in feet,
- \( t \) is the time in seconds,
- the term \( -16t^2 \) represents the effect of gravity (in feet per second squared),
- \( 96t \) represents the initial upward velocity multiplied by time,
- \( 4 \) is the initial height from which the football is kicked.

### Analysis of the Function

1. **Initial Height**:
At \( t = 0 \):
\[
h(0) = -16(0)^2 + 96(0) + 4 = 4 \text{ feet}
\]
This confirms that the football is kicked from a height of 4 feet.

2. **Finding the Time of Maximum Height**:
To find when the football reaches its maximum height, we can use the vertex formula for a quadratic function \( h(t) = at^2 + bt + c \), where the maximum (vertex) occurs at
\[
t = -\frac{b}{2a}
\]
Here, \( a = -16 \) and \( b = 96 \):
\[
t = -\frac{96}{2(-16)} = \frac{96}{32} = 3 \text{ seconds}
\]

3. **Maximum Height**:
We can substitute \( t = 3 \) back into the height equation to find the maximum height:
\[
h(3) = -16(3)^2 + 96(3) + 4
\]
\[
= -16(9) + 288 + 4
\]
\[
= -144 + 288 + 4 = 148 \text{ feet}
\]

4. **Finding When the Football Hits the Ground**:
To find when the football hits the ground, we set \( h(t) = 0 \):
\[
-16t^2 + 96t + 4 = 0
\]
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -16 \), \( b = 96 \), and \( c = 4 \):
\[
t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(4)}}{2(-16)}
\]
\[
t = \frac{-96 \pm \sqrt{9216 + 256}}{-32}
\]
\[
t = \frac{-96 \pm \sqrt{9472}}{-32}
\]
\[
\text{Calculating } \sqrt{9472} \approx 97.35
\]
\[
t = \frac{-96 \pm 97.35}{-32}
\]
Considering the positive root:
\[
t \approx \frac{1.35}{-32} \approx -0.042 \quad (\text{discard this, as time cannot be negative})
\]

For the negative root:
\[
t \approx \frac{-193.35}{-32} \approx 6.05 \text{ seconds}
\]

### Summary
- The maximum height is 148 feet, reached at 3 seconds.
- The football hits the ground approximately 6.06 seconds after being kicked.