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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.

A quadratic function models the height, in feet, of an object above the ground in terms of the time, in seconds, after the object is launched off an elevated surface. The model indicates the object has an initial height of 3 feet above the ground and reaches its maximum height of 403 feet above the ground 5 seconds after being launched. Based on the model, what is the height, in feet, of the object above the ground 6 seconds after being launched?

Learn how to solve a quadratic function problem involving the height of an object launched from an elevated surface. This problem includes concepts such as the vertex form of a quadratic function, initial height, and calculating height at a given time. Suitable for high school students in Grades 9-11.

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