A volleyball is served by a 6-foot player at an initial upward velocity of 33 feet per second. The situation is modeled by the equation h = -16t^2 + 33t + 6h representing the height in feet and t representing the time in seconds. Using this equation, define the domain of the ball when it reaches its maximum height

1.03 seconds
-1.03 seconds
1.22 seconds
23.01 feet

1 answer

To determine the domain of the function that represents the height of the volleyball during its flight, we need to analyze the given equation and the context in which it applies.

The equation provided is:

\[ h(t) = -16t^2 + 33t + 6 \]

Where:

  • \( h(t) \) represents the height of the ball in feet at time \( t \) seconds.
  • The coefficient of \( t^2 \) indicates that the parabola opens downward, which means the ball will reach a maximum point and then fall back down.

Finding the Maximum Height:

To find the time when the ball reaches its maximum height, we can use the formula for the vertex of a parabola given by \( t = -\frac{b}{2a} \):

Here:

  • \( a = -16 \)
  • \( b = 33 \)

Plugging in the values:

\[ t = -\frac{33}{2(-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]

This confirms that the maximum height occurs at approximately 1.03 seconds.

Determining the Domain:

  1. The ball is served at \( t = 0 \) seconds (when it's served), and it will hit the ground (height \( h = 0 \)) at some point after it reaches the maximum height.
  2. We need to find when the ball hits the ground by setting \( h(t) = 0 \):

\[ -16t^2 + 33t + 6 = 0 \]

We can apply the quadratic formula for the above equation:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-33 \pm \sqrt{33^2 - 4(-16)(6)}}{2(-16)} \]

Calculating the discriminant:

\[ b^2 - 4ac = 33^2 - 4(-16)(6) = 1089 + 384 = 1473 \]

Then applying this back in:

\[ t = \frac{-33 \pm \sqrt{1473}}{-32} \]

Calculating \( \sqrt{1473} \approx 38.39 \),

The two possible values of \( t \):

\[ t_1 = \frac{-33 + 38.39}{-32} \quad \text{(will be a negative time, discard)} \] \[ t_2 = \frac{-33 - 38.39}{-32} \approx \frac{-71.39}{-32} \approx 2.23 \text{ seconds} \]

So the time variable \( t \) can range from \( 0 \) seconds to approximately \( 2.23 \) seconds before the ball hits the ground.

Thus the domain of the function regarding the volleyball trajectory can be defined as:

\[ 0 \leq t \leq 2.23 \text{ seconds} \]

To summarize, the maximum height occurs at about 1.03 seconds, and the overall domain of the ball's flight is from 0 seconds to approximately 2.23 seconds before it hits the ground.