To find the domain of the volleyball when it reaches its maximum height, we need to identify when this occurs according to the height equation given:
\[ h(t) = -16t^2 + 33t + 6 \]
This is a quadratic equation in the form of \( h(t) = at^2 + bt + c \), where \( a = -16 \), \( b = 33 \), and \( c = 6 \).
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Find the time \( t \) when the maximum height occurs: The maximum height of a parabola described by \( ax^2 + bx + c \) occurs at \( t = -\frac{b}{2a} \).
\[ t = -\frac{33}{2(-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
This value translates roughly to \( 1.03 \) seconds.
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Determine the height at this time: To find the maximum height, substitute \( t = \frac{33}{32} \) back into the height equation: \[ h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6 \]
However, you are mainly interested in the domain of the volleyball's height at this maximum point.
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Evaluate the domain: Since the volleyball is served upwards from a height of 6 feet, the domain will be from the time it is served ( \( t = 0 \) ) to the time of reaching the ground (when \( h(t) = 0 \)).
To find out when the volleyball hits the ground, set \( h(t) = 0 \): \[ -16t^2 + 33t + 6 = 0 \] Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{-33 \pm \sqrt{33^2 - 4(-16)(6)}}{2(-16)} \] \[ t = \frac{-33 \pm \sqrt{1089 + 384}}{-32} \] \[ t = \frac{-33 \pm \sqrt{1473}}{-32} \] Calculate \( \sqrt{1473} \) and determine the values of \( t \), but you'll find that the relevant time (first root) is when the volleyball returns to height \( h = 0 \) after being served upwards.
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Final conclusion on the domain for the maximum height: Given that we wish to summarize the situation at the maximum height:
The maximum height occurs at approximately \( t = 1.03 \) seconds. Thus, the domain of the volleyball as it reaches maximum height (just before falling back down) involves the time interval from when it is served (\( t = 0 \)) to its maximum height \( t \approx 1.03 \) seconds.
Closed interval for the domain when reaching maximum height: \[ t \in [0, 1.03] \]
In the context of options given, \( 1.03 \) seconds would be the correct choice related to the timing of the peak height.