The equation provided in your homework is:

$h = 112 + 128t - 16t^2$

You are asked to find $h$ for two different values of $t$:

1. $t = 2$
2. $t = 6$

Let's solve for both cases.

a. For $t = 2$:

Substitute $t = 2$ into the equation for $h$:

$h = 112 + 128(2) - 16(2^2)$ $h = 112 + 256 - 16(4)$ $h = 112 + 256 - 64$ $h = 304$

b. For $t = 6$:

Substitute $t = 6$ into the equation for $h$:

$h = 112 + 128(6) - 16(6^2)$ $h = 112 + 768 - 16(36)$ $h = 112 + 768 - 576$ $h = 304$

So for both values $t = 2$ and $t = 6$, $h = 304$.

Would you like more details on this? Here are five related questions you could explore:

1. How can you find the vertex of the quadratic equation $h = 112 + 128t - 16t^2$?
2. What is the significance of the $t^2$ term in the equation?
3. How can you determine when $h$ reaches its maximum value?
4. Can you find when $h = 0$ by solving the quadratic equation?
5. How would you graph this equation and identify its key features?

Tip: When dealing with quadratic equations, remember that they often represent parabolas, and the term with $t^2$ controls the curvature.