Question

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=−16t2+33t+6 h 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 point) Responses −1.03 seconds negative 1.03 seconds 1.22 seconds 1.22 seconds 1.03 seconds 1.03 seconds 23.01 feet

Answer 1

To find the domain of the volleyball when it reaches its maximum height, we need to identify when this occurs according to the height equation given:

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

This is a quadratic equation in the form of \( h(t) = at^2 + bt + c \), where \( a = -16 \), \( b = 33 \), and \( c = 6 \).

1. **Find the time \( t \) when the maximum height occurs:**
The maximum height of a parabola described by \( ax^2 + bx + c \) occurs at \( t = -\frac{b}{2a} \).

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

This value translates roughly to \( 1.03 \) seconds.

2. **Determine the height at this time:**
To find the maximum height, substitute \( t = \frac{33}{32} \) back into the height equation:
\[
h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6
\]

However, you are mainly interested in the domain of the volleyball's height at this maximum point.

3. **Evaluate the domain:**
Since the volleyball is served upwards from a height of 6 feet, the domain will be from the time it is served ( \( t = 0 \) ) to the time of reaching the ground (when \( h(t) = 0 \)).

To find out when the volleyball hits the ground, set \( h(t) = 0 \):
\[
-16t^2 + 33t + 6 = 0
\]
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-33 \pm \sqrt{33^2 - 4(-16)(6)}}{2(-16)}
\]
\[
t = \frac{-33 \pm \sqrt{1089 + 384}}{-32}
\]
\[
t = \frac{-33 \pm \sqrt{1473}}{-32}
\]
Calculate \( \sqrt{1473} \) and determine the values of \( t \), but you'll find that the relevant time (first root) is when the volleyball returns to height \( h = 0 \) after being served upwards.

4. **Final conclusion on the domain for the maximum height:**
Given that we wish to summarize the situation at the maximum height:

The maximum height occurs at approximately \( t = 1.03 \) seconds. Thus, the domain of the volleyball as it reaches maximum height (just before falling back down) involves the time interval from when it is served (\( t = 0 \)) to its maximum height \( t \approx 1.03 \) seconds.

Closed interval for the domain when reaching maximum height:
\[
t \in [0, 1.03]
\]

In the context of options given, \( 1.03 \) seconds would be the correct choice related to the timing of the peak height.