Question
During halftime of a basketball game, a sling shot launches t-shirts into the crowd. The function h=−16t2+32t+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.
Reference Sheet - Features of a Parabola
(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 given the function h(t)=−16t2+32t+5
.
Round to the nearest hundredth, if necessary.
Reference Sheet - Quadratic Formula
(1 point)
A t-shirt will hit the ground at about
seconds.
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.
Reference Sheet - Features of a Parabola
(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 given the function h(t)=−16t2+32t+5
.
Round to the nearest hundredth, if necessary.
Reference Sheet - Quadratic Formula
(1 point)
A t-shirt will hit the ground at about
seconds.
Answers
GPT-4o mini
To determine how long it will take for a t-shirt to reach its maximum height and to find that maximum height, we can use the vertex formula for a parabola.
### Finding the time to reach maximum height
The vertex of a parabola given by the function \( h(t) = at^2 + bt + c \) occurs at the time
\[
t = -\frac{b}{2a}
\]
In the function \( h = -16t^2 + 32t + 5 \), we have:
- \( a = -16 \)
- \( b = 32 \)
Substituting these values into the vertex formula:
\[
t = -\frac{32}{2 \times -16} = -\frac{32}{-32} = 1
\]
So, the t-shirt will reach its maximum height at **1 second**.
### Finding the maximum height
To find the maximum height, we can substitute \( t = 1 \) back into the height function \( h(t) \):
\[
h(1) = -16(1)^2 + 32(1) + 5 = -16 + 32 + 5 = 21
\]
Thus, the maximum height of the t-shirt will be **21 feet**.
### Answer to Question 1
- A t-shirt will reach its maximum height at **1 second**.
- The maximum height of a t-shirt will be **21 feet**.
---
### Question 2 - Finding when the t-shirt hits the ground
To find when the t-shirt hits the ground, we need to find when \( h(t) = 0 \):
\[
-16t^2 + 32t + 5 = 0
\]
We can use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here:
- \( a = -16 \)
- \( b = 32 \)
- \( c = 5 \)
Calculating the discriminant:
\[
b^2 - 4ac = 32^2 - 4(-16)(5) = 1024 + 320 = 1344
\]
Now plug this back into the quadratic formula:
\[
t = \frac{-32 \pm \sqrt{1344}}{2 \times -16} = \frac{-32 \pm \sqrt{1344}}{-32}
\]
Calculating \( \sqrt{1344} \):
\(\sqrt{1344} \approx 36.66\) (rounded to two decimal places).
So plugging this back in:
\[
t = \frac{-32 \pm 36.66}{-32}
\]
Calculating the two possible values for \(t\):
1. (Using the plus sign)
\[
t_1 = \frac{-32 + 36.66}{-32} = \frac{4.66}{-32} \approx -0.15 \quad (\text{not physically possible})
\]
2. (Using the minus sign)
\[
t_2 = \frac{-32 - 36.66}{-32} = \frac{-68.66}{-32} \approx 2.15
\]
Thus, a t-shirt will hit the ground at about **2.15 seconds** (rounded to the nearest hundredth).
### Answer to Question 2
- A t-shirt will hit the ground at about **2.15 seconds**.
### Finding the time to reach maximum height
The vertex of a parabola given by the function \( h(t) = at^2 + bt + c \) occurs at the time
\[
t = -\frac{b}{2a}
\]
In the function \( h = -16t^2 + 32t + 5 \), we have:
- \( a = -16 \)
- \( b = 32 \)
Substituting these values into the vertex formula:
\[
t = -\frac{32}{2 \times -16} = -\frac{32}{-32} = 1
\]
So, the t-shirt will reach its maximum height at **1 second**.
### Finding the maximum height
To find the maximum height, we can substitute \( t = 1 \) back into the height function \( h(t) \):
\[
h(1) = -16(1)^2 + 32(1) + 5 = -16 + 32 + 5 = 21
\]
Thus, the maximum height of the t-shirt will be **21 feet**.
### Answer to Question 1
- A t-shirt will reach its maximum height at **1 second**.
- The maximum height of a t-shirt will be **21 feet**.
---
### Question 2 - Finding when the t-shirt hits the ground
To find when the t-shirt hits the ground, we need to find when \( h(t) = 0 \):
\[
-16t^2 + 32t + 5 = 0
\]
We can use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here:
- \( a = -16 \)
- \( b = 32 \)
- \( c = 5 \)
Calculating the discriminant:
\[
b^2 - 4ac = 32^2 - 4(-16)(5) = 1024 + 320 = 1344
\]
Now plug this back into the quadratic formula:
\[
t = \frac{-32 \pm \sqrt{1344}}{2 \times -16} = \frac{-32 \pm \sqrt{1344}}{-32}
\]
Calculating \( \sqrt{1344} \):
\(\sqrt{1344} \approx 36.66\) (rounded to two decimal places).
So plugging this back in:
\[
t = \frac{-32 \pm 36.66}{-32}
\]
Calculating the two possible values for \(t\):
1. (Using the plus sign)
\[
t_1 = \frac{-32 + 36.66}{-32} = \frac{4.66}{-32} \approx -0.15 \quad (\text{not physically possible})
\]
2. (Using the minus sign)
\[
t_2 = \frac{-32 - 36.66}{-32} = \frac{-68.66}{-32} \approx 2.15
\]
Thus, a t-shirt will hit the ground at about **2.15 seconds** (rounded to the nearest hundredth).
### Answer to Question 2
- A t-shirt will hit the ground at about **2.15 seconds**.