Question

Jesse tossed a paint brush off her roof. The height of the brush (in meters above the ground)
\[t\] seconds after Jesse tossed it is modeled by
\[h(t)=-5t^2+5t+10\]
Jesse wants to know when the brush will hit the ground.
1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation.
\[h(t)=\]
2) How many seconds after being thrown does the brush hit the ground?
seconds

Answer 1

To rewrite the height function \( h(t) = -5t^2 + 5t + 10 \) in a different form, we will complete the square.

1) **Completing the Square:**

First, factor out \(-5\) from the first two terms:

\[
h(t) = -5(t^2 - t) + 10
\]

Next, take half of the coefficient of \(t\) (which is \(-1\)), square it, and add and subtract this value inside the parentheses. Half of \(-1\) is \(-\frac{1}{2}\), and squaring it gives \(\frac{1}{4}\).

\[
h(t) = -5\left(t^2 - t + \frac{1}{4} - \frac{1}{4}\right) + 10
\]

Now rewrite it as:

\[
h(t) = -5\left((t - \frac{1}{2})^2 - \frac{1}{4}\right) + 10
\]

Distributing \(-5\):

\[
h(t) = -5(t - \frac{1}{2})^2 + \frac{5}{4} + 10
\]

Convert \(10\) to quarters:

\[
10 = \frac{40}{4}
\]

Adding the constants gives:

\[
h(t) = -5(t - \frac{1}{2})^2 + \frac{45}{4}
\]

Thus, the vertex form of the function is:

\[
h(t) = -5\left(t - \frac{1}{2}\right)^2 + \frac{45}{4}
\]

2) **Finding when the brush hits the ground:**

To find when the brush hits the ground, we set \(h(t) = 0\):

\[
-5(t - \frac{1}{2})^2 + \frac{45}{4} = 0
\]

Rearranging gives:

\[
-5(t - \frac{1}{2})^2 = -\frac{45}{4}
\]

Dividing both sides by \(-5\):

\[
(t - \frac{1}{2})^2 = \frac{45}{20} = \frac{9}{4}
\]

Now, take the square root of both sides:

\[
t - \frac{1}{2} = \pm\frac{3}{2}
\]

Solving for \(t\):

1. \(t - \frac{1}{2} = \frac{3}{2}\):

\[
t = 2
\]

2. \(t - \frac{1}{2} = -\frac{3}{2}\):

\[
t = -1 \quad (\text{not a valid solution for time})
\]

Thus, the brush hits the ground **2 seconds** after being thrown.

**Final Answer:**
1) \(h(t) = -5\left(t - \frac{1}{2}\right)^2 + \frac{45}{4}\)

2) The brush hits the ground after \(2\) seconds.