1. Ms. Whitmell goes on a trip to the moon and jumps from a hill into a crater.Her height with respect to time is modeled by the equation h(t)=-4t^2+12t+7 where t is time in minutes and h(t) is height in metres. a. What is the height of the hill? b. How long is Ms Whitmell in the air? c. What is Ms Whitmell's maximum height and at what time does this happen?

## Step 1: Finding the initial height<br />### The height of the hill is the height at $t=0$. Substituting $t=0$ into the equation $h(t) = -4t^2 + 12t + 7$ gives $h(0) = -4(0)^2 + 12(0) + 7 = 7$.<br />## Step 2: Finding the time in the air<br />### Ms. Whitmell lands when $h(t) = 0$. We solve $-4t^2 + 12t + 7 = 0$ for $t$. Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $t = \frac{-12 \pm \sqrt{12^2 - 4(-4)(7)}}{2(-4)} = \frac{-12 \pm \sqrt{144 + 112}}{-8} = \frac{-12 \pm \sqrt{256}}{-8} = \frac{-12 \pm 16}{-8}$. This gives two solutions: $t = \frac{-12 + 16}{-8} = \frac{4}{-8} = -\frac{1}{2}$ and $t = \frac{-12 - 16}{-8} = \frac{-28}{-8} = \frac{7}{2}$. Since time cannot be negative, the time in the air is $\frac{7}{2}$ minutes, or 3.5 minutes.<br />## Step 3: Finding the maximum height and time<br />### The maximum height occurs at the vertex of the parabola. The $t$-coordinate of the vertex is given by $t = -\frac{b}{2a} = -\frac{12}{2(-4)} = \frac{12}{8} = \frac{3}{2} = 1.5$ minutes. Substituting $t = \frac{3}{2}$ into the height equation: $h(\frac{3}{2}) = -4(\frac{3}{2})^2 + 12(\frac{3}{2}) + 7 = -4(\frac{9}{4}) + 18 + 7 = -9 + 18 + 7 = 16$ meters.