To analyze the motion of the volleyball described by the equation \( h(t) = -16t^2 + 33t + 6 \), where \( h \) represents height in feet and \( t \) represents time in seconds, we can break down the components of the equation and find important characteristics such as maximum height, time to reach the maximum height, and when the volleyball will hit the ground.
1. Maximum Height
The equation is a quadratic function in the standard form \( h(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 33 \)
- \( c = 6 \)
The time \( t \) at which the maximum height occurs can be calculated using the formula: \[ t = -\frac{b}{2a} \]
Substituting in the values: \[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
Next, we substitute this \( t \) value back into the height equation to find the maximum height: \[ h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6 \]
Calculating it step-by-step:
- Calculate \( \left( \frac{33}{32} \right)^2 = \frac{1089}{1024} \)
- Calculate \( -16 \times \frac{1089}{1024} = -\frac{17424}{1024} \)
- Calculate \( 33 \times \frac{33}{32} = \frac{1089}{32} = \frac{34752}{1024} \)
- Now combine: \[ h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{34752}{1024} + 6 \]
- Convert \( 6 \) to a fraction: \( 6 = \frac{6144}{1024} \)
- Combine: \[ h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{34752}{1024} + \frac{6144}{1024} \] \[ = \frac{-17424 + 34752 + 6144}{1024} = \frac{23472}{1024} \approx 22.921875 \text{ feet} \]
2. Time When the Volleyball Hits the Ground
To find when the volleyball hits the ground, we need to solve \( h(t) = 0 \): \[ -16t^2 + 33t + 6 = 0 \]
Using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting the values: \[ t = \frac{-33 \pm \sqrt{33^2 - 4 \cdot -16 \cdot 6}}{2 \cdot -16} \] Calculating the discriminant: \[ 33^2 = 1089 \] \[ 4 \cdot 16 \cdot 6 = 384 \] So, \[ t = \frac{-33 \pm \sqrt{1089 + 384}}{-32} = \frac{-33 \pm \sqrt{1473}}{-32} \] Calculating \( \sqrt{1473} \) gives approximately \( 38.4 \).
Now substituting: \[ t = \frac{-33 \pm 38.4}{-32} \]
This yields two potential times:
- \( t = \frac{-33 + 38.4}{-32} \approx -0.165625 \) (not valid since time can't be negative)
- \( t = \frac{-33 - 38.4}{-32} \approx \frac{-71.4}{-32} \approx 2.228125 \)
Summary
- The volleyball reaches its maximum height of approximately 22.92 feet at about 1.03 seconds.
- The volleyball hits the ground approximately 2.23 seconds after being served.