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The given height equation is \( h(t) = -16t^2 + 96t + 640 \), which represents the height of the rocket in feet as a function of time \( t \) seconds.

### 1. **Maximum Height of the Rocket:**

To find the maximum height, we need to determine the vertex of the quadratic equation. The time \( t \) at which the maximum height occurs is given by the formula for the vertex of a parabola:

\[
t = \frac{-b}{2a}
\]

Here, \( a = -16 \), \( b = 96 \), and \( c = 640 \) (though \( c \) is not needed for this step).

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

Now, substitute \( t = 3 \) back into the original equation to find the maximum height:

\[
h(3) = -16(3)^2 + 96(3) + 640
\]
\[
h(3) = -16(9) + 288 + 640 = -144 + 288 + 640 = 784 \text{ feet}
\]

So, the rocket's maximum height is **784 feet**.

### 2. **Time to Reach Maximum Height:**

We have already calculated that the rocket reaches its maximum height at **3 seconds**.

### 3. **Time to Reach the Ground:**

To find when the rocket reaches the ground, we set \( h(t) = 0 \) and solve for \( t \):

\[
-16t^2 + 96t + 640 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting the values of \( a = -16 \), \( b = 96 \), and \( c = 640 \):

\[
t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(640)}}{2(-16)}
\]
\[
t = \frac{-96 \pm \sqrt{9216 + 40960}}{-32}
\]
\[
t = \frac{-96 \pm \sqrt{50176}}{-32}
\]
\[
t = \frac{-96 \pm 224}{-32}
\]

Now, calculate the two possible values for \( t \):

\[
t_1 = \frac{-96 + 224}{-32} = \frac{128}{-32} = -4 \text{ seconds (not possible, discard this value)}
\]
\[
t_2 = \frac{-96 - 224}{-32} = \frac{-320}{-32} = 10 \text{ seconds}
\]

So, the rocket reaches the ground after **10 seconds**.

---

### Summary of Answers:
- The rocket's maximum height is **784 feet**.
- It takes **3 seconds** to reach the maximum height.
- The rocket reaches the ground after **10 seconds**.

Would you like further details or have any questions?

---

### Follow-up Questions:
1. What is the initial height of the rocket when \( t = 0 \)?
2. How fast is the rocket moving when it hits the ground?
3. What is the velocity of the rocket at its maximum height?
4. Can you calculate the time when the rocket is halfway to its maximum height?
5. How would the equation change if the rocket was launched with a different initial velocity?

### Tip:
To determine the velocity of an object from its height function, take the derivative of the height function with respect to time.

A rocket is fired upward from some initial distance above the ground. Its height (in feet), h, above the ground t seconds after it is fired is given by h(t) = -16t^2 + 96t + 640. What is the rocket's maximum height? How long does it take for the rocket to reach its maximum height? After it is fired, the rocket reaches the ground at t = ? seconds.

This problem involves solving a quadratic equation to determine the maximum height of a rocket, the time it takes to reach the maximum height, and when the rocket hits the ground. Using algebraic concepts like the vertex formula and quadratic formula, the problem is ideal for students in Grades 9-11.

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