The problem involves a quadratic function representing the height of a rocket over time, where the equation given is:

$h(t) = -4.9t^2 + 139t + 423$

To find the maximum height, we need to determine the vertex of this quadratic function because it represents a parabolic curve that opens downward (since the coefficient of $t^2$ is negative). The time at which the maximum height is reached can be found using the vertex formula:

$t = \frac{-b}{2a}$

Here, $a = -4.9$, $b = 139$, and $c = 423$ (though we don't need $c$ for the vertex).

First, calculate the time at which the rocket reaches its peak:

$t = \frac{-139}{2(-4.9)} = \frac{-139}{-9.8} = 14.18 \, \text{seconds} \, (\text{rounded to 2 decimal places})$

Now, substitute this value of $t$ back into the equation for $h(t)$ to find the height at the peak:

$h(14.18) = -4.9(14.18)^2 + 139(14.18) + 423$

$h(14.18) = -4.9(201.09) + 1969.02 + 423$

$h(14.18) = -985.34 + 1969.02 + 423 = 1406.68 \, \text{meters}$

Thus, the rocket peaks at approximately 1406.68 meters above sea level.

Would you like any further clarification or details on this?

Here are 5 related questions for further practice:

1. How do you find the time of flight for the rocket using the quadratic equation?
2. What does the discriminant of the quadratic function tell you about the motion of the rocket?
3. How would the maximum height change if the coefficient of $t^2$ were larger in magnitude?
4. What is the initial velocity of the rocket based on the given equation?
5. How long does it take for the rocket to return to sea level after being launched?

Tip: When working with quadratic functions that model motion, always check if the parabola opens upwards or downwards by examining the sign of the $t^2$-term.