Question

each year, the cider hills town council entertains the town’s residents with fireworks. this year, they hired someone to launch fireworks from the top of the town’s tallest building, which is 432 feet tall. the first firework that is launched reaches a maximum height of 1,008 feet above the ground at 6 seconds after launch.
write the height of the firework above the ground in feet, h(t), as a quadratic function of the time in seconds, t, after the firework is launched. write the equation for the function in the form h(t) = at² + bt + c.
simplify any fractions.
h(t) =

Explanation:

Step1: Identify known values

We know the following:

• The initial height (when \( t = 0 \)) is the height of the building, so \( h(0)=432 \) feet.
• The time to reach maximum height for a quadratic function \( h(t)=at^{2}+bt + c \) (which is a parabola opening downwards since it's a projectile motion) is at \( t = 3 \) seconds (assuming the "3 seconds after launch" from the problem, maybe a typo in the original text, but we'll proceed with \( t = 3 \) as the time to max height) and the maximum height \( h(3)=1008 \) feet. Also, for a quadratic function \( h(t)=at^{2}+bt + c \), the vertex (maximum point for \( a<0 \)) occurs at \( t=-\frac{b}{2a} \).

Step2: Use \( t = 0 \) to find \( c \)

When \( t = 0 \), \( h(0)=a(0)^{2}+b(0)+c=c \). Since \( h(0) = 432 \), we have \( c = 432 \). So the function is \( h(t)=at^{2}+bt + 432 \).

Step3: Use the vertex time and height

The vertex is at \( t = 3 \), so \( -\frac{b}{2a}=3 \), which implies \( b=-6a \). Also, when \( t = 3 \), \( h(3)=a(3)^{2}+b(3)+432=9a + 3b+432 \). We know \( h(3) = 1008 \), so substitute \( b=-6a \) into this equation:

$$
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$$

Step4: Find \( b \)

Since \( b=-6a \) and \( a=-64 \), then \( b=-6\times(-64) = 384 \).

Step5: Write the function

Now we have \( a=-64 \), \( b = 384 \), and \( c = 432 \). So the quadratic function is:

$$ h(t)=-64t^{2}+384t + 432 $$

Answer:

\( h(t)=-64t^{2}+384t + 432 \)