Question
At baseball practice, Matt hit the ball into the outfield. The function h(t)=−16t2+32t+3 gives the height h(t) of the baseball in feet, t seconds after it is hit. A) What is the value of h(1) ? Response area B) What does h(1.74) represent in context? Response area C) Which value would not be valid in the context of this problem?
Answers
GPT-4o mini
To address the questions based on the function \( h(t) = -16t^2 + 32t + 3 \), let's solve each part step by step.
### A) What is the value of h(1)?
To find \( h(1) \), we substitute \( t = 1 \) into the function:
\[
h(1) = -16(1)^2 + 32(1) + 3
\]
\[
= -16(1) + 32 + 3
\]
\[
= -16 + 32 + 3
\]
\[
= 16 + 3
\]
\[
= 19
\]
So, **the value of \( h(1) \) is 19 feet**.
### B) What does h(1.74) represent in context?
The value \( h(1.74) \) represents the height of the baseball, in feet, at \( t = 1.74 \) seconds after it has been hit.
### C) Which value would not be valid in the context of this problem?
The height function \( h(t) \) models the height of the baseball in feet. Since the height cannot be negative, any value of \( t \) that yields a negative height in \( h(t) \) would not be valid in this context.
To determine the values of \( t \) for which \( h(t) < 0 \):
1. Set the function to zero to find the time when the ball hits the ground:
\[
-16t^2 + 32t + 3 = 0
\]
2. Applying the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 32 \), and \( c = 3 \):
\[
t = \frac{-32 \pm \sqrt{32^2 - 4(-16)(3)}}{2(-16)}
\]
\[
= \frac{-32 \pm \sqrt{1024 + 192}}{-32}
\]
\[
= \frac{-32 \pm \sqrt{1216}}{-32}
\]
\[
= \frac{-32 \pm 34}{-32}
\]
3. Calculating the two potential outcomes:
- \( t_1 = \frac{-32 + 34}{-32} = \frac{2}{-32} = -\frac{1}{16} \) (not valid since time cannot be negative)
- \( t_2 = \frac{-32 - 34}{-32} = \frac{-66}{-32} = 2.0625 \) (valid, as it represents the time when the ball reaches the ground)
Thus, any \( t \) greater than approximately \( 2.06 \) seconds would lead to a negative height, and therefore would not be valid. Therefore, valid values of \( t \) are \( t \geq 0 \) and \( t \leq 2.0625 \).
### Summary:
- A) \( h(1) = 19 \) feet.
- B) \( h(1.74) \) represents the height of the baseball 1.74 seconds after it was hit.
- C) Any \( t > 2.0625 \) seconds would not be valid as it would lead to a negative height.
### A) What is the value of h(1)?
To find \( h(1) \), we substitute \( t = 1 \) into the function:
\[
h(1) = -16(1)^2 + 32(1) + 3
\]
\[
= -16(1) + 32 + 3
\]
\[
= -16 + 32 + 3
\]
\[
= 16 + 3
\]
\[
= 19
\]
So, **the value of \( h(1) \) is 19 feet**.
### B) What does h(1.74) represent in context?
The value \( h(1.74) \) represents the height of the baseball, in feet, at \( t = 1.74 \) seconds after it has been hit.
### C) Which value would not be valid in the context of this problem?
The height function \( h(t) \) models the height of the baseball in feet. Since the height cannot be negative, any value of \( t \) that yields a negative height in \( h(t) \) would not be valid in this context.
To determine the values of \( t \) for which \( h(t) < 0 \):
1. Set the function to zero to find the time when the ball hits the ground:
\[
-16t^2 + 32t + 3 = 0
\]
2. Applying the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 32 \), and \( c = 3 \):
\[
t = \frac{-32 \pm \sqrt{32^2 - 4(-16)(3)}}{2(-16)}
\]
\[
= \frac{-32 \pm \sqrt{1024 + 192}}{-32}
\]
\[
= \frac{-32 \pm \sqrt{1216}}{-32}
\]
\[
= \frac{-32 \pm 34}{-32}
\]
3. Calculating the two potential outcomes:
- \( t_1 = \frac{-32 + 34}{-32} = \frac{2}{-32} = -\frac{1}{16} \) (not valid since time cannot be negative)
- \( t_2 = \frac{-32 - 34}{-32} = \frac{-66}{-32} = 2.0625 \) (valid, as it represents the time when the ball reaches the ground)
Thus, any \( t \) greater than approximately \( 2.06 \) seconds would lead to a negative height, and therefore would not be valid. Therefore, valid values of \( t \) are \( t \geq 0 \) and \( t \leq 2.0625 \).
### Summary:
- A) \( h(1) = 19 \) feet.
- B) \( h(1.74) \) represents the height of the baseball 1.74 seconds after it was hit.
- C) Any \( t > 2.0625 \) seconds would not be valid as it would lead to a negative height.