To determine the maximum height of an object launched with the function $h(t) = -16t^2 + 360t$, we need to find the vertex of this parabolic equation, since the vertex will give us the maximum height (as the coefficient of $t^2$ is negative, indicating a parabola that opens downwards).

The vertex form of a parabola $ax^2 + bx + c$ is given by the vertex $(h, k)$, where $h = -\frac{b}{2a}$. In our case, the function is:

$h(t) = -16t^2 + 360t$

Here, $a = -16$ and $b = 360$.

First, find the time $t$ at which the maximum height occurs:

$t = -\frac{b}{2a} = -\frac{360}{2 \cdot -16} = \frac{360}{32} = 11.25$

Now, substitute $t = 11.25$ back into the function $h(t)$ to find the maximum height:

$h(11.25) = -16(11.25)^2 + 360(11.25)$

Calculate $(11.25)^2$:

$(11.25)^2 = 126.5625$

Now substitute and calculate:

$h(11.25) = -16 \cdot 126.5625 + 360 \cdot 11.25$ $h(11.25) = -2025 + 4050$ $h(11.25) = 2025$

Therefore, the maximum height of the object is $2025$ feet.

Would you like more details or have any questions?

Here are 5 questions you might consider asking:

1. How do you determine the vertex of a parabola?
2. Can you explain why the vertex represents the maximum height in this context?
3. What are the units used in this problem?
4. How does the coefficient of $t^2$ affect the direction of the parabola?
5. Can this method be used for any quadratic function?

Tip: Always check the direction of the parabola (whether it opens upwards or downwards) to determine if you're finding a maximum or minimum value.