We are given the height function of the rocket as:

$h(t) = -4.9t^2 + 208t + 367$

To solve this, we need to answer two questions:

1. Time of splashdown (when the rocket hits the ocean):

Splashdown occurs when the height $h(t)$ is 0. To find the time of splashdown, we solve the equation:

$h(t) = 0 \implies -4.9t^2 + 208t + 367 = 0$

This is a quadratic equation in the form $at^2 + bt + c = 0$, where $a = -4.9$, $b = 208$, and $c = 367$. We'll use the quadratic formula to find $t$:

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

2. Height at the peak:

The peak of the rocket's flight occurs at the vertex of the parabola. For a quadratic equation $ax^2 + bx + c$, the time at which the peak occurs is given by:

$t_{\text{peak}} = \frac{-b}{2a}$

At that time, we can substitute the value of $t_{\text{peak}}$ into the height function to find the maximum height.

Let's solve both of these step by step.

Step 1: Time of splashdown (solving for $t$) and height at the peak.### Results:

1. Splashdown time: The rocket splashes down into the ocean after approximately $44.15$ seconds.
2. Maximum height: The rocket reaches its peak height at around $2574.35$ meters above sea level.

Do you need further details or have any additional questions?

Here are 5 related questions to expand on this topic:

1. How would the splashdown time change if the initial velocity were doubled?
2. What is the significance of the coefficient $-4.9$ in the height equation?
3. Can you calculate the rocket's velocity at the time it reaches the peak?
4. How would you determine the total distance traveled by the rocket during its flight?
5. How would you modify the equation if the rocket started from an elevated platform?

Tip: In quadratic motion problems, the coefficient of the squared term represents the acceleration due to gravity, especially when dealing with heights in meters and time in seconds.