(1 point) NASA launches a rocket at t=0
 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t2+277t+395
.

Assuming that the rocket will splash down into the ocean, after how many seconds does splashdown occur?
Round your answer to the nearest hundredth.
The rocket splashes down after 
57.96
functionsequation editor seconds.

How high above sea-level does the rocket get at its peak? Round your answer to the nearest hundredth.

The rocket peaks at 
functionsequation editor meters above sea-level.

We are given the height function of the rocket as:

\[
h(t) = -4.9t^2 + 277t + 395
\]

### 1. Time of Splashdown:
To find the time when the rocket splashes down, we set \( h(t) = 0 \) (since it will be at sea level when it splashes down).

\[
-4.9t^2 + 277t + 395 = 0
\]

We solve this quadratic equation using the quadratic formula:

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

Where:
- \(a = -4.9\)
- \(b = 277\)
- \(c = 395\)

Substitute the values into the quadratic formula:

\[
t = \frac{-277 \pm \sqrt{277^2 - 4(-4.9)(395)}}{2(-4.9)}
\]
\[
t = \frac{-277 \pm \sqrt{76729 - (-7746)}}{-9.8}
\]
\[
t = \frac{-277 \pm \sqrt{84475}}{-9.8}
\]
\[
t = \frac{-277 \pm 290.66}{-9.8}
\]

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

1. \(t_1 = \frac{-277 + 290.66}{-9.8} = \frac{13.66}{-9.8} \approx -1.396\) (not physically meaningful as time cannot be negative).
2. \(t_2 = \frac{-277 - 290.66}{-9.8} = \frac{-567.66}{-9.8} \approx 57.96\)

Thus, the rocket splashes down at **57.96 seconds**.

### 2. Maximum Height (Peak):
The maximum height occurs at the vertex of the parabola, which is given by the formula for the time at the vertex of a parabola:

\[
t_{\text{vertex}} = \frac{-b}{2a}
\]

Substitute the values of \(a = -4.9\) and \(b = 277\):

\[
t_{\text{vertex}} = \frac{-277}{2(-4.9)} = \frac{-277}{-9.8} \approx 28.3 \text{ seconds}
\]

Now, substitute \(t = 28.3\) into the height equation to find the maximum height:

\[
h(28.3) = -4.9(28.3)^2 + 277(28.3) + 395
\]
\[
h(28.3) = -4.9(801.29) + 277(28.3) + 395
\]
\[
h(28.3) = -3936.34 + 7830.11 + 395
\]
\[
h(28.3) \approx 4288.77 \text{ meters}
\]

Thus, the rocket reaches a maximum height of **4288.77 meters** above sea level.

---

### Final Answers:
- The rocket splashes down after **57.96 seconds**.
- The rocket reaches a peak height of **4288.77 meters**.

Would you like more details or have any questions on this? 

### Further Questions:
1. How does the coefficient \( -4.9 \) affect the rocket's trajectory?
2. How do you calculate the maximum height of a parabola?
3. What is the significance of the negative time solution in the quadratic equation?
4. How would you graph this quadratic function to visualize the rocket's path?
5. What would happen if the value of \( b \) were increased?

**Tip**: When solving quadratic equations, always check both possible solutions, but discard those that don't make sense in the context (e.g., negative time in this case).

NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t²+277t+395. Assuming that the rocket will splash down into the ocean, after how many seconds does splashdown occur? Round your answer to the nearest hundredth. The rocket splashes down after 57.96 seconds. How high above sea-level does the rocket get at its peak? Round your answer to the nearest hundredth.

Learn how to calculate the splashdown time and peak height of a rocket using quadratic equations. This problem involves solving a quadratic equation for splashdown and finding the vertex to determine the rocket's maximum height. Suitable for high school students (Grades 9-11).

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