To find the domain of the volleyball's height function when it reaches its maximum height, we first need to identify the time at which the ball reaches this maximum height. The given height equation is:
\[ h(t) = -16t^2 + 33t + 6 \]
This is a quadratic function, and the maximum height can be found using the vertex formula \( t = -\frac{b}{2a} \), where \( a = -16 \) and \( b = 33 \).
Calculating this gives:
\[ t = -\frac{33}{2(-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
This value represents the time when the volleyball reaches its maximum height.
Next, since we need the domain during which the ball is in the air, we will examine the height function \( h(t) \) until it returns to the height of zero after being served. Setting the height function to zero gives us the time when the ball hits the ground:
\[ -16t^2 + 33t + 6 = 0 \]
We can solve this equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where:
- \( a = -16 \)
- \( b = 33 \)
- \( c = 6 \)
Calculating the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 33^2 - 4(-16)(6) = 1089 + 384 = 1473 \]
Now applying the quadratic formula:
\[ t = \frac{-33 \pm \sqrt{1473}}{2(-16)} \]
Calculating \( \sqrt{1473} \) (approximately 38.38):
\[ t = \frac{-33 \pm 38.38}{-32} \]
Calculating the two possible values for \( t \):
- \( t = \frac{-33 + 38.38}{-32} \approx -0.17 \) (Not valid in this context)
- \( t = \frac{-33 - 38.38}{-32} \)
Calculating the second option:
\[ t \approx \frac{-71.38}{-32} \approx 2.23 \text{ seconds} \]
Thus, the ball reaches the maximum height at approximately \( 1.03 \) seconds and lands back on the ground at approximately \( 2.23 \) seconds.
Therefore, the domain of the ball when it reaches its maximum height is:
\[ [0, 2.23] \]
The specific answer options related to the maximum height include \( 1.03 \) seconds, which is when the ball reaches its maximum height. So, the answer from the provided options that defines this time is:
1.03 seconds.