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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:

23.01 feet

1.03 seconds

−1.03 seconds

1.22 seconds

9 answers

To determine the time when the volleyball reaches its maximum height, we can use the formula for the vertex of a quadratic function given by \(h(t) = -16t^2 + 33t + 6\). The time at which the maximum height occurs can be found using the formula:

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

where \(a = -16\) and \(b = 33\) in our equation.

Plugging in these values, we have:

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

To define the domain of the ball when it reaches its maximum height, we note that the volleyball is served at \(t = 0\) and will rise until it reaches its peak height at \(t \approx 1.03\) seconds. After this point, the height will begin to decrease until it reaches the ground.

Therefore, the domain is:

\[ [0, 1.03] \]

From the options provided, the maximum height occurs at approximately \(1.03\) seconds, which is the correct answer.

Final Answer: 1.03 seconds

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=−4.9t^2+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?
Responses:

10

5

1/10

2

To determine when the rock hits the water, set the height equation \( h = -4.9t^2 + 10 \) equal to 0 (since the height is 0 when it hits the water):

\[ -4.9t^2 + 10 = 0 \] \[ 4.9t^2 = 10 \] \[ t^2 = \frac{10}{4.9} \approx 2.0408 \] \[ t \approx \sqrt{2.0408} \approx 1.43 \text{ seconds} \]

The rock hits the water approximately 1.43 seconds after it is thrown.

To set up a reasonable scale for the range (which in this case refers to the time values in seconds that you'd record while determining when the rock hits the water), you would want to have increments that adequately cover the time before the rock hits the water.

Looking at the options provided:

  1. 10 seconds - This would cover much more than needed.
  2. 5 seconds - This is still more than required, as the rock hits the water in approximately 1.43 seconds.
  3. 1/10 seconds - This would be very precise, allowing for fine increments, but may be unnecessarily detailed for this context.
  4. 2 seconds - This is just enough to cover the necessary time and slightly more for ease of documenting (allowing for a stopping point after the rock hits).

Given the options, a reasonable scale would be 2 seconds, as it provides a straightforward and manageable range while covering the time it takes for the rock to hit the water.

Final Answer: 2

An owl swoops down from its perch in a tree at 20 feet to catch field mice below. It follows the function h=−6t^2−24t+20 . Plot points to graph the equation. Which ordered pair represents the vertex?(1 point)
Responses

(44,−2)

(−2,44)

(1,0)

(−5,0)

To find the vertex of the quadratic function \( h = -6t^2 - 24t + 20 \), we can use the vertex formula. The vertex \((t, h)\) of a quadratic function \(h = at^2 + bt + c\) can be found using:

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

Here, \(a = -6\) and \(b = -24\).

Step 1: Calculate \(t\)

\[ t = -\frac{-24}{2 \cdot -6} = \frac{24}{-12} = -2 \]

Step 2: Calculate \(h\)

Now, plug \(t = -2\) back into the function to find the corresponding height \(h\):

\[ h = -6(-2)^2 - 24(-2) + 20 \] \[ h = -6(4) + 48 + 20 \] \[ h = -24 + 48 + 20 \] \[ h = 44 \]

Vertex Coordinates

Thus, the vertex is at the point:

\[ (-2, 44) \]

Conclusion

The ordered pair representing the vertex of the function \( h = -6t^2 - 24t + 20 \) is:

Final Answer: \((-2, 44)\)

Luca is building a garden in his backyard. He needs to erect a fence to keep his dog Luna from digging in it. He has enough materials for a fence with 120 feet around the perimeter. This function is represented by the formula A=−w2+60w , where A is the area and w is the width. Which of the following correctly identifies the x-intercept(s)?
Responses:

(0,0)

(0,0) and (60,0)

(60,0)

(60,0) and (30,900)

To find the x-intercept(s) of the quadratic function \( A = -w^2 + 60w \), we need to set the area \( A \) equal to zero and solve for \( w \):

\[ -w^2 + 60w = 0 \]

We can factor this equation:

\[ w(-w + 60) = 0 \]

Setting each factor equal to zero gives us:

  1. \( w = 0 \)
  2. \( -w + 60 = 0 \) ⟹ \( w = 60 \)

Thus, the x-intercepts (where \( A = 0 \)) are at the points \( (0, 0) \) and \( (60, 0) \).

Summary of X-intercepts

  • The x-intercepts of the function are \((0, 0)\) and \((60, 0)\).

Final Answer: (0,0) and (60,0)

An owl swoops down from its perch in a tree at 30 feet to catch field mice. It follows the function h=6t2 −24t+30 , where t is the time in seconds and h is the height in feet. Which of the following best interprets the vertex?
Responses:

(30,0)

(6,2)

(2,6)

(0,30)

To interpret the vertex of the quadratic function \( h = 6t^2 - 24t + 30 \), we can find the vertex using the vertex formula for a quadratic function in the form \( h = at^2 + bt + c \), where:

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

Step 1: Identify \(a\) and \(b\)

In the function \( h = 6t^2 - 24t + 30 \):

  • \( a = 6 \)
  • \( b = -24 \)

Step 2: Calculate \(t\)

Now, plug in the values for \( b \) and \( a \):

\[ t = -\frac{-24}{2 \cdot 6} = \frac{24}{12} = 2 \]

Step 3: Calculate \(h\)

Now, we plug \( t = 2 \) back into the function to find the corresponding height \( h \):

\[ h = 6(2)^2 - 24(2) + 30 \] \[ h = 6(4) - 48 + 30 \] \[ h = 24 - 48 + 30 \] \[ h = 6 \]

Vertex Coordinates

Thus, the vertex of the function is at the point \( (2, 6) \).

Interpretation of the Vertex

In this context, the vertex \( (2, 6) \) represents the point in time (2 seconds) when the owl reaches its maximum height of 6 feet.

Final Answer: (2, 6)