Luca Is In A Culvert Below Street Level. He Launches An Object At An Upward Velocity Of 40 Feet Per Second. Use GeoGebra To Graph The Situation Using The Formula $y=-16t^2+40t-5$. How Far Above Or Below Street Level Is Luca's Initial Launching

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Luca's Launch: A Mathematical Exploration of Projectile Motion

Projectile motion is a fundamental concept in physics that describes the trajectory of an object under the influence of gravity. In this article, we will explore the motion of an object launched from a culvert below street level using the formula y=−16t2+40t−5y=-16t^2+40t-5. We will use GeoGebra to visualize the situation and determine how far above or below street level Luca's initial launching point is.

The formula y=−16t2+40t−5y=-16t^2+40t-5 represents the height of the object above the ground as a function of time. The coefficients of the formula have the following meanings:

  • −16-16 is the acceleration due to gravity (in feet per second squared)
  • 4040 is the initial upward velocity (in feet per second)
  • −5-5 is the initial height of the object above the ground (in feet)

To visualize the situation, we can use GeoGebra to graph the formula y=−16t2+40t−5y=-16t^2+40t-5. We can create a graph with time on the x-axis and height on the y-axis. The graph will show the trajectory of the object as a parabola.

**GeoGebra Graph**
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![GeoGebra Graph](https://example.com/geogebra-graph.png)

From the graph, we can see that the object is launched from a point below street level. The parabola opens downward, indicating that the object is under the influence of gravity. The vertex of the parabola represents the maximum height reached by the object.

To find the initial launching point, we need to determine the time at which the object is launched. We can do this by setting the height of the object to zero and solving for time.

**Solving for Time**
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$0 = -16t^2 + 40t - 5$

$16t^2 - 40t + 5 = 0$

$t = \frac{40 \pm \sqrt{(-40)^2 - 4(16)(5)}}{2(16)}$

$t = \frac{40 \pm \sqrt{1600 - 320}}{32}$

$t = \frac{40 \pm \sqrt{1280}}{32}$

$t = \frac{40 \pm 35.78}{32}$

$t = \frac{40 + 35.78}{32}$ or $t = \frac{40 - 35.78}{32}$

$t = 2.23$ or $t = 0.11$

The time at which the object is launched is t=0.11t = 0.11 seconds.

Now that we have the time at which the object is launched, we can find the initial height by plugging this value into the formula.

**Finding the Initial Height**
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$y = -16(0.11)^2 + 40(0.11) - 5$

$y = -1.66 + 4.4 - 5$

$y = -2.26$

The initial height of the object above the ground is −2.26-2.26 feet.

In this article, we used GeoGebra to visualize the situation of an object launched from a culvert below street level. We analyzed the graph and found the initial launching point and the initial height of the object above the ground. The initial height of the object above the ground is −2.26-2.26 feet, indicating that Luca's initial launching point is below street level.

In our previous article, we explored the motion of an object launched from a culvert below street level using the formula y=−16t2+40t−5y=-16t^2+40t-5. We used GeoGebra to visualize the situation and determined how far above or below street level Luca's initial launching point is. In this article, we will answer some frequently asked questions related to the topic.

Q: What is the significance of the formula y=−16t2+40t−5y=-16t^2+40t-5?

A: The formula y=−16t2+40t−5y=-16t^2+40t-5 represents the height of the object above the ground as a function of time. The coefficients of the formula have the following meanings:

  • −16-16 is the acceleration due to gravity (in feet per second squared)
  • 4040 is the initial upward velocity (in feet per second)
  • −5-5 is the initial height of the object above the ground (in feet)

Q: How do you graph the situation using GeoGebra?

A: To graph the situation using GeoGebra, you can create a graph with time on the x-axis and height on the y-axis. The graph will show the trajectory of the object as a parabola.

Q: What is the time at which the object is launched?

A: The time at which the object is launched is t=0.11t = 0.11 seconds.

Q: What is the initial height of the object above the ground?

A: The initial height of the object above the ground is −2.26-2.26 feet.

Q: Why is the initial height negative?

A: The initial height is negative because the object is launched from a point below street level.

Q: What is the significance of the vertex of the parabola?

A: The vertex of the parabola represents the maximum height reached by the object.

Q: How do you find the time at which the object reaches its maximum height?

A: To find the time at which the object reaches its maximum height, you can use the formula t=−b2at = \frac{-b}{2a}, where aa and bb are the coefficients of the quadratic equation.

Q: What is the maximum height reached by the object?

A: The maximum height reached by the object is y=4.4y = 4.4 feet.

Q: How do you find the time at which the object hits the ground?

A: To find the time at which the object hits the ground, you can set the height of the object to zero and solve for time.

Q: What is the time at which the object hits the ground?

A: The time at which the object hits the ground is t=1.25t = 1.25 seconds.

In this article, we answered some frequently asked questions related to the topic of Luca's launch. We hope that this article has provided you with a better understanding of the motion of an object launched from a culvert below street level.