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+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)

Question

Responses

1.22 seconds
1.22 seconds

23.01 feet
23.01 feet

1.03 seconds
1.03 seconds

−1.03 seconds

Answer 1

To find the time at which the volleyball reaches its maximum height, we can analyze the equation given:

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

This is a quadratic equation, and the maximum height occurs at the vertex of the parabola. The time at which the vertex occurs can be found using the formula:

\[ t = -\frac{b}{2a} \]

In this equation, \( a \) is the coefficient of \( t^2 \) (which is -16), and \( b \) is the coefficient of \( t \) (which is 33).

Plugging in the values:

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

Rounding this value, the time at which the volleyball reaches its maximum height is approximately 1.03 seconds.

The domain of the function for the time until the ball hits the ground is from \( t = 0 \) to the time it takes to hit the ground. However, since you're specifically asking for the time at maximum height, we conclude:

The time to reach maximum height is approximately 1.03 seconds.

Therefore, the correct response is: 1.03 seconds.

Answer 2

While hiking, Marek throws a rock off a cliff that is roughly 10 meters above a lake. This situation is modeled by the equation h=−4.9t2+10 , with h representing the height in meters and t the time in seconds. If you were to set up a table to determine when the rock hits the water, what would be a reasonable scale for the range?(1 point)
Responses

110
1 tenth

2
2

10
10

5

Answer 3

To determine when the rock hits the water, we need to find out when the height \( h \) of the rock equals 0, which represents the rock hitting the surface of the lake.

The equation modeling the height of the rock is:

\[ h = -4.9t^2 + 10 \]

To find the time \( t \) when the rock hits the water, we set \( h = 0 \):

\[ 0 = -4.9t^2 + 10 \]

Rearranging this gives:

\[ 4.9t^2 = 10 \]

Now divide both sides by 4.9:

\[ t^2 = \frac{10}{4.9} \]

Now calculate:

\[ t^2 \approx 2.0408 \]

Taking the square root gives:

\[ t \approx \sqrt{2.0408} \approx 1.43 \text{ seconds} \]

Since the rock hits the water at approximately 1.43 seconds, a reasonable range for the time scale would go from 0 seconds to just above 1.43 seconds.

Looking at the provided options:

1. 110: This is far too high.
2. 1 tenth: This would cover a very short duration.
3. 2: A range from 0 to 2 seconds is reasonable, as it allows for the rock to hit the water after approximately 1.43 seconds.
4. 10: This is excessive.
5. 5: This is also excessive.

Therefore, the most reasonable scale for the range, considering that the rock hits the water at approximately 1.43 seconds, is:

The correct response is: 2.