For this question we can think about what we know:

At the starting time, the balloon is released at 45 cubits in height

After 2 seconds it reaches a maximum of 55 cubits off the ground

**PART A)** We want to write a formula h(t) to represent the
height od the balloon at a given time called t. There are different equations
for quadratics, one being standard form of y=ax^2 + bx + c, and one being
vertex form, or y = a(x-h)^2 + k. Because we know the maximum height and time
it took to get there, we have the information to plug in for the vertex (h,k) =
(2, 55). Let's plug that into the vertex form formula, but we'll use t and h(t)
instead of x and y:

h(t) = a(t-2)^2 + 55

Now we want to fill in whatever else we know so that we can finish our equation. We have the information of one other point in the balloon's path, which is the starting point: at a time = 0 seconds, the balloon is 45 cubits up. Let's plug this in for our time and height:

45 = a(0-2)^2 + 55

Now the only variable in our equation is a, so we can use al of this information to solve for a:

45 = a * (-2)^2 + 55

45 = a * 4 + 55

-10 = a* 4

-5/2 = a

Now let's plug this back into our vertex form equation:

h(t) = (-5/2)(t-2)^2 + 55

**a) The formula for the height of the balloon at some time t
is h(t) = (-5/2)(t-2)^2 + 55**

**
PART B)** To find the time the balloon hits the ground, let's think
about what that means... At a certain time t that we want to find, the balloon
with be at what height? Our formula tells us the height of the balloon above
the ground, so when the balloon hits the ground, it is at a height of 0. To
find the time when it hits the ground, all we have to do is plug in 0 as our
height and solve for the time t.

0 = (-5/2)(t-2)^2 + 55 Subtract 55

-55 = (-5/2)(t-2)^2 Divide by -5/2

22 = (t - 2)^2 Square root both sides

sqrt(22) = t - 2 Add 2

t = sqrt(22) + 2 = 6.69041..

**b) The balloon will hit the ground after 6.69 seconds.**

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.