To solve this problem, we can model the height $h(t)$ of the object as a quadratic function of time $t$, given by:

$h(t) = at^2 + bt + c$

where:

• $c$ is the initial height of the object (3 feet),
• the object reaches a maximum height of 403 feet at $t = 5$ seconds.

Step 1: Set up the function

Since the height reaches its maximum at $t = 5$, we know that the vertex of the parabola occurs at $t = 5$. The form of a quadratic function that reaches its maximum (or minimum) at $t = h$ can be expressed as:

$h(t) = a(t - h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

In this case:

• The vertex $(h, k) = (5, 403)$,
• The initial height $h(0) = 3$.

Thus, we can write the equation as:

$h(t) = a(t - 5)^2 + 403$

Step 2: Find $a$ using the initial height

We know $h(0) = 3$. Substitute $t = 0$ and $h(0) = 3$ into the equation:

$3 = a(0 - 5)^2 + 403$

This simplifies to:

$3 = 25a + 403$

Solving for $a$:

$25a = 3 - 403$ $25a = -400$ $a = -16$

Step 3: Write the final function

Now we have the function:

$h(t) = -16(t - 5)^2 + 403$

Step 4: Calculate $h(6)$

Substitute $t = 6$ into the function to find the height 6 seconds after launch:

$h(6) = -16(6 - 5)^2 + 403$ $h(6) = -16(1)^2 + 403$ $h(6) = -16 + 403$ $h(6) = 387$

Answer

The height of the object above the ground 6 seconds after being launched is 387 feet.

Would you like further details or have any questions?

Here are five related questions to deepen understanding:

1. How would the equation change if the maximum height was reached at a different time?
2. What if the initial height was different? How would that impact the function?
3. Can you derive the same result using a different method, such as completing the square?
4. How would this model look if the object reached its maximum height at the ground level?
5. What real-world scenarios could require modifying the parameters of this quadratic function?

Tip: When working with parabolas, knowing the vertex form $h(t) = a(t - h)^2 + k$ can simplify finding maximum or minimum points.