Question
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=-16^2+40t-5
a 0 feet above or below
b 2.5 feet below
c 5 feet below
d 5 feet above
a 0 feet above or below
b 2.5 feet below
c 5 feet below
d 5 feet above
Answers
Answered by
GPT-4o mini
To graph the situation using the quadratic equation for the motion of the object, we need to correctly interpret and represent the formula \( y = -16t^2 + 40t - 5 \). This equation models the height \( y \) of the object above or below ground level (0 feet) as a function of time \( t \) in seconds.
Where:
- The term \(-16t^2\) indicates the acceleration due to gravity (using feet/seconds).
- \(40t\) is the initial upward velocity (in feet per second).
- The constant term \(-5\) indicates that the launching point of the object is 5 feet below street level.
### Steps to Graph the Situation:
1. **Set up the function**: The equation \( y = -16t^2 + 40t - 5 \) represents a parabola opening downward.
2. **Identify the vertex**: The vertex of the parabola gives you the maximum height reached by the object. The time \( t \) at which this occurs can be found using the formula \( t = -\frac{b}{2a} \) where \( a = -16 \) and \( b = 40 \).
\[
t = -\frac{40}{2(-16)} = \frac{40}{32} = 1.25 \text{ seconds}
\]
3. **Calculate the maximum height**: Substitute \( t = 1.25 \) back into the equation:
\[
y = -16(1.25)^2 + 40(1.25) - 5
\]
\[
y = -16(1.5625) + 50 - 5
\]
\[
y = -25 + 50 - 5 = 20 \text{ feet}
\]
This means that at 1.25 seconds, the object reaches a maximum height of 20 feet above street level (which is 15 feet above the launch point since he launches from 5 feet below ground).
4. **Graph**: Using GeoGebra or another graphing tool, plot the function \( y = -16t^2 + 40t - 5 \) over a reasonable time interval, say from \( t = 0 \) to \( t = 3 \) seconds, to observe the object's trajectory.
### Analyze Heights
- **0 feet above or below (point A)**: At \( t = 0 \):
\[
y = -5 \text{ feet (5 feet below)}
\]
- **2.5 feet below (point B)**: Set \( y = -2.5 \) to find when the object is 2.5 feet below street level:
\[
-2.5 = -16t^2 + 40t - 5
\]
Rearranging gives:
\[
0 = -16t^2 + 40t - 2.5
\]
Use the quadratic formula to solve for \( t \).
- **5 feet below (point C)**: Set \( y = -5 \):
\[
0 = -16t^2 + 40t - 5
\]
Again, this is a quadratic equation to solve.
- **5 feet above (point D)**: Set \( y = 5 \):
\[
5 = -16t^2 + 40t - 5
\]
Which rearranges to:
\[
0 = -16t^2 + 40t - 10
\]
Another quadratic to solve.
### Final Notes
To execute these calculations and precise graphing, you might want to utilize GeoGebra software directly for plotting the graph and finding the roots of the quadratic equations. By investigating the roots, you can observe at which points the object is at the specific heights (A, B, C, and D) specified. Remember that the time and heights are indeed key variables in this quadratic motion context!
Where:
- The term \(-16t^2\) indicates the acceleration due to gravity (using feet/seconds).
- \(40t\) is the initial upward velocity (in feet per second).
- The constant term \(-5\) indicates that the launching point of the object is 5 feet below street level.
### Steps to Graph the Situation:
1. **Set up the function**: The equation \( y = -16t^2 + 40t - 5 \) represents a parabola opening downward.
2. **Identify the vertex**: The vertex of the parabola gives you the maximum height reached by the object. The time \( t \) at which this occurs can be found using the formula \( t = -\frac{b}{2a} \) where \( a = -16 \) and \( b = 40 \).
\[
t = -\frac{40}{2(-16)} = \frac{40}{32} = 1.25 \text{ seconds}
\]
3. **Calculate the maximum height**: Substitute \( t = 1.25 \) back into the equation:
\[
y = -16(1.25)^2 + 40(1.25) - 5
\]
\[
y = -16(1.5625) + 50 - 5
\]
\[
y = -25 + 50 - 5 = 20 \text{ feet}
\]
This means that at 1.25 seconds, the object reaches a maximum height of 20 feet above street level (which is 15 feet above the launch point since he launches from 5 feet below ground).
4. **Graph**: Using GeoGebra or another graphing tool, plot the function \( y = -16t^2 + 40t - 5 \) over a reasonable time interval, say from \( t = 0 \) to \( t = 3 \) seconds, to observe the object's trajectory.
### Analyze Heights
- **0 feet above or below (point A)**: At \( t = 0 \):
\[
y = -5 \text{ feet (5 feet below)}
\]
- **2.5 feet below (point B)**: Set \( y = -2.5 \) to find when the object is 2.5 feet below street level:
\[
-2.5 = -16t^2 + 40t - 5
\]
Rearranging gives:
\[
0 = -16t^2 + 40t - 2.5
\]
Use the quadratic formula to solve for \( t \).
- **5 feet below (point C)**: Set \( y = -5 \):
\[
0 = -16t^2 + 40t - 5
\]
Again, this is a quadratic equation to solve.
- **5 feet above (point D)**: Set \( y = 5 \):
\[
5 = -16t^2 + 40t - 5
\]
Which rearranges to:
\[
0 = -16t^2 + 40t - 10
\]
Another quadratic to solve.
### Final Notes
To execute these calculations and precise graphing, you might want to utilize GeoGebra software directly for plotting the graph and finding the roots of the quadratic equations. By investigating the roots, you can observe at which points the object is at the specific heights (A, B, C, and D) specified. Remember that the time and heights are indeed key variables in this quadratic motion context!