The equation for the height $h(t)$ of the rocket as a function of time is:

$h(t) = -16t^2 + 240t + 1216$

We need to find the maximum height of the rocket. For a quadratic function $h(t) = at^2 + bt + c$, the maximum height occurs at the vertex, which can be found using the formula:

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

Here, $a = -16$ and $b = 240$. Substituting these values:

$t_{\text{max}} = -\frac{240}{2(-16)} = \frac{240}{32} = 7.5 \, \text{seconds}.$

Next, substitute $t_{\text{max}} = 7.5$ into the height equation to find the maximum height:

$h(7.5) = -16(7.5)^2 + 240(7.5) + 1216$

First, calculate $(7.5)^2$:

$(7.5)^2 = 56.25$

Now substitute:

$h(7.5) = -16(56.25) + 240(7.5) + 1216$

$h(7.5) = -900 + 1800 + 1216 = 2116 \, \text{feet}.$

Thus, the maximum height is:

$\boxed{2116 \, \text{feet}}$

Would you like a breakdown of any specific steps, or help with related questions?

Related Questions:

1. How long does it take for the rocket to hit the ground after being launched?
2. What is the initial velocity of the rocket when it is launched?
3. How would the maximum height change if the initial velocity were increased?
4. At what times is the rocket at a height of 1500 feet?
5. How do changes in the coefficient of $t^2$ affect the trajectory of the rocket?

Tip:

For quadratic equations describing motion, the vertex formula $t = -b/(2a)$ is key to finding the time at which the maximum or minimum occurs.