Question

During halftime of a basketball game, a sling shot launches t-shirts into the crowd. The function h=−16t2+72t+5 models the t-shirt's height h in feet after t seconds.

Use the vertex formula to determine how many seconds it will take a t-shirt to reach the maximum height. Then find the maximum height of the t-shirt.

Round to the hundredths place, if necessary.

(2 points)
A t-shirt will reach its maximum height at
seconds.

The maximum height of a t-shirt will be
feet.

Question 2
Use the quadratic formula to calculate the time it will take for a t-shirt shot outside the stands to hit the ground. Round to the nearest hundredth, if necessary.(1 point)
A t-shirt will hit the ground at about
seconds.

Answer 1

To find the time \( t \) when the t-shirt reaches its maximum height, we can use the vertex formula for a quadratic function, which is given by:

\[
t = -\frac{b}{2a}
\]

In the function \( h = -16t^2 + 72t + 5 \):

- \( a = -16 \)
- \( b = 72 \)

Plugging in the values, we get:

\[
t = -\frac{72}{2 \times -16} = -\frac{72}{-32} = \frac{72}{32} = 2.25
\]

So, a t-shirt will reach its maximum height at **2.25 seconds**.

Now, to find the maximum height \( h \), we substitute \( t = 2.25 \) back into the height equation:

\[
h = -16(2.25)^2 + 72(2.25) + 5
\]

Calculating \( (2.25)^2 \):

\[
(2.25)^2 = 5.0625
\]

Then substituting:

\[
h = -16(5.0625) + 72(2.25) + 5
\]
\[
= -81 + 162 + 5
\]
\[
= 81 + 5 = 86
\]

So, the maximum height of a t-shirt will be **86 feet**.

### Summary of the First Question:
- A t-shirt will reach its maximum height at **2.25 seconds**.
- The maximum height of a t-shirt will be **86 feet**.

### Question 2: Calculating when the t-shirt hits the ground

To find out when the t-shirt hits the ground, we need to set \( h = 0 \) and solve for \( t \):

\[
0 = -16t^2 + 72t + 5
\]

We can use the quadratic formula to solve for \( t \):

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting \( a = -16 \), \( b = 72 \), and \( c = 5 \):

\[
t = \frac{-72 \pm \sqrt{(72)^2 - 4(-16)(5)}}{2(-16)}
\]
\[
= \frac{-72 \pm \sqrt{5184 + 320}}{-32}
\]
\[
= \frac{-72 \pm \sqrt{5504}}{-32}
\]
\[
= \frac{-72 \pm 74.33}{-32} \quad \text{(approximately, since } \sqrt{5504} \approx 74.33\text{)}
\]

Now calculate the two possibilities:

1. \( t_1 = \frac{-72 + 74.33}{-32} \)
2. \( t_2 = \frac{-72 - 74.33}{-32} \)

Calculating \( t_2 \) (as it will give us the valid time when it hits the ground):

\[
t_2 = \frac{-146.33}{-32} \approx 4.57
\]

Thus, a t-shirt will hit the ground at about **4.57 seconds** (rounded to the nearest hundredth).

### Summary of the Second Question:
- A t-shirt will hit the ground at about **4.57 seconds**.