To determine the maximum height of a volleyball in the air, we use the projectile motion equation:
\[ h(t) = -16t^2 + vt + h_0 \]
where:
- \( v = 33 \) feet per second (initial speed),
- \( h_0 = 6 \) feet (initial height).
Plugging in these values gives:
\[ h(t) = -16t^2 + 33t + 6 \]
To find the time \( t \) when the volleyball reaches its highest point, we use the vertex formula:
\[ t = -\frac{b}{2a} \]
Here, \( a = -16 \) and \( b = 33 \):
\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \text{ seconds} \approx 1.03 \text{ seconds} \]
Next, we calculate the height at this time:
\[ h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6 \]
Calculating \( \left(\frac{33}{32}\right)^2 = \frac{1089}{1024} \):
Now substituting into the height equation:
- \(-16 \cdot \frac{1089}{1024} = -\frac{17424}{1024}\)
- \(33 \cdot \frac{33}{32} = \frac{1089}{32} = \frac{34656}{1024}\)
- \(6 = \frac{6144}{1024}\)
Putting it all together:
\[ h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{34656}{1024} + \frac{6144}{1024} = \frac{6144}{1024} = 6 \]
Thus, the volleyball reaches a maximum height of 6 feet at approximately 1.03 seconds after being thrown.