https://file.aiquickdraw.com/mathbot/file/1729744688660dhgtf3oy.jpg

Let's break down the problem from the image:

The height of the rocket is modeled by the quadratic equation:
\[
h(t) = -4.9t^2 + 1200t + 450
\]
Where:
- \( h(t) \) is the height in meters after \( t \) seconds,
- \( t \) is the time in seconds.

### a) From what height was the rocket launched?
The height at launch means the height when \( t = 0 \). To find this, we substitute \( t = 0 \) into the equation:

\[
h(0) = -4.9(0)^2 + 1200(0) + 450 = 450 \text{ meters}.
\]

So, the rocket was launched from a height of 450 meters.

### b) What is the maximum height the rocket reaches?
Since the equation is quadratic with a negative coefficient for \( t^2 \), it opens downward, meaning there is a maximum point (the vertex of the parabola). The time at which the maximum height occurs is given by the formula:

\[
t = \frac{-b}{2a}
\]
where \( a = -4.9 \) and \( b = 1200 \).

\[
t = \frac{-1200}{2(-4.9)} = \frac{-1200}{-9.8} = 122.45 \text{ seconds}.
\]

Now, substitute \( t = 122.45 \) back into the height equation to find the maximum height:

\[
h(122.45) = -4.9(122.45)^2 + 1200(122.45) + 450.
\]

Let me calculate this for you:

\[
h(122.45) \approx -4.9(14989.3) + 146940 + 450 \approx -73446 + 146940 + 450 = 73944 \text{ meters}.
\]

So, the maximum height the rocket reaches is approximately 73,944 meters.

### c) When will the rocket splash down?
To find when the rocket splashes down (i.e., when it hits the ocean), we need to solve for \( t \) when \( h(t) = 0 \):

\[
-4.9t^2 + 1200t + 450 = 0.
\]

We can solve this quadratic equation using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = -4.9 \), \( b = 1200 \), and \( c = 450 \).

Let me calculate this for you:

\[
t = \frac{-1200 \pm \sqrt{1200^2 - 4(-4.9)(450)}}{2(-4.9)}.
\]

This simplifies to:

\[
t = \frac{-1200 \pm \sqrt{1440000 + 8820}}{-9.8},
\]
\[
t = \frac{-1200 \pm \sqrt{1448820}}{-9.8},
\]
\[
t = \frac{-1200 \pm 1203.67}{-9.8}.
\]

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

\[
t_1 = \frac{-1200 + 1203.67}{-9.8} \approx \frac{3.67}{-9.8} \approx -0.37 \text{ (ignore negative time)},
\]
\[
t_2 = \frac{-1200 - 1203.67}{-9.8} \approx \frac{-2403.67}{-9.8} \approx 245.27 \text{ seconds}.
\]

So, the rocket will splash down after approximately 245.27 seconds.

---

Would you like further details or clarification on any of these steps? Here are 5 related questions you might consider:

1. How would the rocket's launch height change if the constant 450 were different?
2. How does the coefficient of \( t^2 \) affect the shape of the rocket's trajectory?
3. How could you find the time the rocket takes to reach half its maximum height?
4. What does the discriminant tell us about the nature of the rocket's flight time?
5. How would air resistance affect the height equation?

**Tip:** When solving quadratic equations for real-world problems, always consider whether negative solutions make sense in the given context (e.g., negative time).

A rocket is launched, and its height above sea level t seconds after launch is given by the equation h(t) = -4.9t^2 + 1200t + 450. a) From what height was the rocket launched? b) What is the maximum height the rocket reaches? c) If the rocket will splash down in the ocean, when will it splash down?

Explore the rocket's trajectory given by the quadratic function h(t) = -4.9t^2 + 1200t + 450. Learn how to calculate the launch height, maximum height, and splashdown time by solving quadratic equations. Suitable for high school students (Grades 9-12).

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