To determine the domain of the volleyball when it reaches its maximum height using the given equation \( h = -16t^2 + 33t + 6 \), we first need to find the time at which the maximum height occurs.
The function \( h(t) = -16t^2 + 33t + 6 \) is a quadratic function, and it opens downward (since the coefficient of \( t^2 \) is negative). The maximum height occurs at the vertex of this parabola, which can be found using the formula for the vertex \( t = -\frac{b}{2a} \), where \( a = -16 \) and \( b = 33 \).
Substituting the values:
\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \approx 1.03125 , \text{seconds} \]
This result indicates that the maximum height occurs around \( 1.03 \) seconds.
Now we need to define the domain of the problem, which consists of times from when the volleyball is served (at \( t = 0 \) seconds) until it reaches the ground again (when \( h = 0 \)).
To find when the ball reaches the ground, we set the height \( h \) to zero:
\[ 0 = -16t^2 + 33t + 6 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = -16 \)
- \( b = 33 \)
- \( c = 6 \)
Calculating the discriminant:
\[ b^2 - 4ac = 33^2 - 4(-16)(6) = 1089 + 384 = 1473 \]
Now substituting into the quadratic formula:
\[ t = \frac{-33 \pm \sqrt{1473}}{2 \times -16} \]
Calculating \( \sqrt{1473} \approx 38.4 \):
\[ t = \frac{-33 \pm 38.4}{-32} \]
Calculating the two potential values for \( t \):
- \( t = \frac{-33 + 38.4}{-32} \) (this will give a negative time, which we discard)
- \( t = \frac{-33 - 38.4}{-32} \approx \frac{71.4}{32} \approx 2.23 , \text{seconds} \)
Thus, the ball is served from \( t = 0 \) seconds up to approximately \( t = 2.23 \) seconds. Therefore, the domain of \( t \) when the ball reaches its maximum height (rounded to two decimal places) is:
From 0 to approximately 1.03 seconds.
So the relevant answer reflecting the time it reaches maximum height is:
1.03 seconds
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