To find the domain of the ball when it reaches its maximum height, we first need to determine when the maximum height occurs in the projectile's motion described by the equation:
\[ h(t) = -16t^2 + 33t + 6 \]
This is a quadratic equation, and it reaches its maximum at the vertex. The time \( t \) at which the vertex occurs for a quadratic function of the form \( at^2 + bt + c \) can be found using the formula:
\[ t = -\frac{b}{2a} \]
Here, \( a = -16 \) and \( b = 33 \).
Let's calculate:
\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \]
This means that the maximum height occurs at \( t = \frac{33}{32} \) seconds.
Next, we need to define the domain of the ball's height when it reaches its maximum height. Since the ball is served at an initial height of 6 feet and experiences both ascent and descent, we also need to find when the ball hits the ground (height \( h = 0 \)). To find this, we can set the height equation to zero and solve for \( t \):
\[ -16t^2 + 33t + 6 = 0 \]
We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting \( a = -16 \), \( b = 33 \), and \( c = 6 \):
\[ t = \frac{-33 \pm \sqrt{33^2 - 4 \cdot (-16) \cdot 6}}{2 \cdot (-16)} \] \[ t = \frac{-33 \pm \sqrt{1089 + 384}}{-32} \] \[ t = \frac{-33 \pm \sqrt{1473}}{-32} \]
Simplifying \( \sqrt{1473} \):
\[ \sqrt{1473} \approx 38.4 \]
So,
\[ t = \frac{-33 \pm 38.4}{-32} \]
Calculating the two potential times:
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For the plus sign: \[ t = \frac{5.4}{-32} \approx -0.16875 \quad \text{(not valid, as time cannot be negative)} \]
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For the negative sign: \[ t = \frac{-71.4}{-32} \approx 2.22625 \quad \text{(valid, as time must be positive)} \]
Thus, the ball is in the air from \( t = 0 \) seconds (when served) to \( t \approx 2.23 \) seconds (when it hits the ground). Therefore, the domain of the ball when it reaches its maximum height is:
\[ [0, \frac{33}{32}] \]
Thus, the answer is:
Domain: \( [0, \frac{33}{32}] \) seconds.