To model the height of the volleyball over time, you can use the equation of a projectile motion. The general form of the height \( h \) as a function of time \( t \) can be given by the formula:
\[ h(t) = -16t^2 + vt + h_0 \]
where:
- \( v \) is the initial velocity (in feet per second),
- \( h_0 \) is the initial height (in feet),
- and the coefficient \(-16\) represents the effect of gravity in feet per second squared.
Given:
- Initial height \( h_0 = 6 \) feet
- Initial velocity \( v = 33 \) feet per second
Substituting these values into the equation:
\[ h(t) = -16t^2 + 33t + 6 \]
To find the vertex of the parabola represented by this equation, we can use the formula for the vertex \( t = -\frac{b}{2a} \) where \( a \) and \( b \) are the coefficients from the standard form \( at^2 + bt + c \).
Here:
- \( a = -16 \)
- \( b = 33 \)
Now substituting these values:
\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \text{ seconds} \]
Now we can find 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 \( \left(\frac{33}{32}\right)^2 \):
\[ \left(\frac{33}{32}\right)^2 = \frac{1089}{1024} \]
Substituting back into the height equation:
\[ h\left(\frac{33}{32}\right) = -16 \cdot \frac{1089}{1024} + 33 \cdot \frac{33}{32} + 6 \] \[ h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{1089}{32} + 6 \] \[ = -\frac{17424}{1024} + \frac{34656}{1024} + \frac{6144}{1024} \] \[ = \frac{17424 - 17424 + 6144}{1024} \] \[ = \frac{6144}{1024} = 6 \]
As a result, the vertex of the function is:
\[ \left(\frac{33}{32}, 6\right) \]
In conclusion, the vertex of the function representing the height of the volleyball is at:
\[ \left(\frac{33}{32}, 6\right) \approx (1.03, 6) \text{ (in feet)} \]
This means the maximum height reached is 6 feet at approximately \( t = 1.03 \) seconds.
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