To find the maximum height reached by the volleyball and define the domain for when it reaches that height, we start with the given equation for height \( h \):
\[ h(t) = -16t^2 + 33t + 6 \]
This is a quadratic equation of the form \( h(t) = at^2 + bt + c \), where \( a = -16 \), \( b = 33 \), and \( c = 6 \).
Step 1: Find the time at which the maximum height occurs.
The maximum height of a quadratic function occurs at the vertex. The time \( t \) at which the vertex is located can be found using the formula:
\[ t = -\frac{b}{2a} \]
Substituting in the values of \( b \) and \( a \):
\[ t = -\frac{33}{2 \cdot (-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
Step 2: Calculate the maximum height.
Next, substitute \( t \) back into the height equation to find the maximum height \( h(t) \):
\[ h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6 \]
Calculating \( \left(\frac{33}{32}\right)^2 \) and substituting:
\[ \left(\frac{33}{32}\right)^2 = \frac{1089}{1024} \] \[ h\left(\frac{33}{32}\right) = -16 \cdot \frac{1089}{1024} + 33 \cdot \frac{33}{32} + 6 \]
Calculating each term:
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The first term: \[ -16 \cdot \frac{1089}{1024} = -\frac{17424}{1024} = -17.0078125 \]
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The second term: \[ 33 \cdot \frac{33}{32} = \frac{1089}{32} = 34.03125 \]
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The third term: \[ 6 = \frac{192}{32} = 6 \]
Now combine these values: \[ h\left(\frac{33}{32}\right) = -17.0078125 + 34.03125 + 6 \]
Calculating this gives: \[ h\left(\frac{33}{32}\right) = 23.0234375 \text{ feet} \approx 23 \text{ feet} \]
Domain of the Ball
To specify the domain of the ball during which it reaches its maximum height, we must also consider when the ball will return to the ground (when \( h(t) = 0 \)).
To find that time, solve for \( t \) in the equation:
\[ -16t^2 + 33t + 6 = 0 \]
Use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting in \( a = -16, b = 33, c = 6 \):
\[ t = \frac{-33 \pm \sqrt{33^2 - 4 \cdot (-16) \cdot 6}}{2 \cdot (-16)} \] \[ = \frac{-33 \pm \sqrt{1089 + 384}}{-32} \] \[ = \frac{-33 \pm \sqrt{1473}}{-32} \]
Calculating \( \sqrt{1473} \approx 38.4 \):
\[ t \approx \frac{-33 \pm 38.4}{-32} \]
Calculating the two possible values for \( t \):
- \( t = \frac{-33 + 38.4}{-32} \)
- \( t = \frac{-33 - 38.4}{-32} \)
Solving these gives:
- \( t_1 = \frac{5.4}{-32} \) (ignore because it's negative)
- \( t_2 = \frac{-71.4}{-32} \approx 2.23 \text{ seconds} \)
Conclusion
The maximum height occurs at approximately \( t = 1.03125 \) seconds, while the ball will hit the ground approximately at \( t = 2.23 \) seconds. Therefore, the domain of the ball is:
\[ [0, 2.23] \text{ seconds} \]
The maximum height of the volleyball is approximately 23.0234375 feet.