Question

At the county July 4th celebration, a firework is launched. The function h(t) = -16t^2 + 96t gives the height in feet of the firework, t seconds after it has been launched.

Answer 1

The function \( h(t) = -16t^2 + 96t \) models the height, in feet, of a firework \( t \) seconds after it has been launched. To analyze this function, we can identify several key characteristics:

1. **Vertex of the Parabola**: The function is a quadratic equation in standard form \( h(t) = at^2 + bt + c \), where \( a = -16 \), \( b = 96 \), and \( c = 0 \). The vertex of the parabola can be found using the formula \( t = -\frac{b}{2a} \).

\[
t = -\frac{96}{2 \times -16} = -\frac{96}{-32} = 3 \text{ seconds}
\]

To find the maximum height, we substitute \( t = 3 \) back into the height function:

\[
h(3) = -16(3^2) + 96(3)
\]
\[
h(3) = -16(9) + 288
\]
\[
h(3) = -144 + 288 = 144 \text{ feet}
\]

Thus, the maximum height of the firework is **144 feet**, reached at **3 seconds** after launch.

2. **Finding the Times When the Firework Hits the Ground**: To determine when the firework hits the ground, we set \( h(t) = 0 \) and solve for \( t \):

\[
-16t^2 + 96t = 0
\]

Factoring out common terms:

\[
-16t(t - 6) = 0
\]

Setting each factor equal to zero gives:

\[
t = 0 \quad \text{or} \quad t - 6 = 0 \implies t = 6
\]

Therefore, the firework is launched at \( t = 0 \) seconds and hits the ground again at \( t = 6 \) seconds.

In summary:

- The firework reaches a maximum height of **144 feet** at **3 seconds** after launch.
- It hits the ground again at **6 seconds** after launch.