To find the time when the volleyball reaches its maximum height, we can use the vertex formula for a quadratic equation of the form \( h = at^2 + bt + c \). The time at which the maximum height occurs can be found using the formula:
\[ t = -\frac{b}{2a} \]
In this equation, \( a = -16 \) and \( b = 33 \).
Substituting the values in gives:
\[ t = -\frac{33}{2(-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \approx 1.03 \text{ seconds} \]
Now, regarding the domain, the volleyball will be served at \( t = 0 \) seconds, and it will reach the ground when the height \( h \) is equal to 0. To determine when the ball hits the ground, we can solve for \( t \) when \( h = 0 \):
\[ -16t^2 + 33t + 6 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16, b = 33, c = 6 \):
- Calculate the discriminant:
\[ b^2 - 4ac = 33^2 - 4(-16)(6) = 1089 + 384 = 1473 \]
- Now apply the quadratic formula:
\[ t = \frac{-33 \pm \sqrt{1473}}{2 \times -16} \] \[ t \approx \frac{-33 \pm 38.41}{-32} \]
Calculating the two possible times:
- Using the positive root:
\[ t = \frac{-33 + 38.41}{-32} \approx \frac{5.41}{-32} \approx -0.169 \text{ (not valid since time cannot be negative)} \]
- Using the negative root:
\[ t = \frac{-33 - 38.41}{-32} \approx \frac{-71.41}{-32} \approx 2.23 \text{ seconds} \]
Now, the domain of the volleyball is from \( t = 0 \) to \( t \approx 2.23 \) seconds.
Thus, the ball reaches its maximum height at approximately 1.03 seconds. This means the correct response from the options you provided is:
1.03 seconds.