To find the domain of the volleyball's height in relation to time until it reaches its maximum height, we can analyze the quadratic equation given:
\[ h = -16t^2 + 33t + 6 \]
This equation represents a parabola that opens downward (since the coefficient of \(t^2\) is negative). The maximum height occurs at the vertex of the parabola.
The time \(t\) at which the maximum height occurs can be calculated using the vertex formula \(t = -\frac{b}{2a}\), where \(a = -16\) and \(b = 33\).
Calculating \(t\):
\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
Next, we need to determine the time when the ball hits the ground. The height \(h\) will be equal to zero when the ball reaches the ground. We solve the equation \(h = 0\):
\[ 0 = -16t^2 + 33t + 6 \]
To solve this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we have:
- \(a = -16\)
- \(b = 33\)
- \(c = 6\)
First, we calculate the discriminant:
\[ b^2 - 4ac = 33^2 - 4(-16)(6) = 1089 + 384 = 1473 \]
Now, applying the quadratic formula:
\[ t = \frac{-33 \pm \sqrt{1473}}{2(-16)} = \frac{-33 \pm \sqrt{1473}}{-32} \]
Calculating \(\sqrt{1473} \approx 38.4\):
Thus, we have:
\[ t = \frac{-33 + 38.4}{-32} \quad \text{and} \quad t = \frac{-33 - 38.4}{-32} \]
Calculating these values:
- For \(t = \frac{5.4}{-32} \approx -0.16875\) (discard this, as time cannot be negative)
- For \(t = \frac{-71.4}{-32} \approx 2.22625\)
Therefore, the ball will hit the ground at approximately \(t \approx 2.226\) seconds.
To determine the domain of the ball's height as it rises to the maximum height and then falls back to the ground, we can say the domain is:
\[ [t_0, t_1] = [0, \approx 2.226] \]
Thus, the domain of the ball when it reaches its maximum height and falls back to the ground is:
\[ [0, 2.226] \] seconds.
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