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=−16t2+40t−5. How far above or below street level is Luca’s initial launching point?(1 point) Responses
0 feet above or below
2.5 feet below
5 feet above
5 feet below
To graph the situation using the formula y = -16t^2 + 40t - 5 in GeoGebra, follow these steps:
1. Open GeoGebra and create a new 2D graphing window.
2. Enter the equation y = -16t^2 + 40t - 5 into the input bar at the top of the window.
3. Press Enter to graph the equation.
4. Adjust the viewing window to see the graph clearly.
Based on the graph of the equation, Luca's initial launching point is 2.5 feet below street level.
Huh??
When t = 0, y = -5
Therefore the launch is 5 feet below the street where y presumably is zero.
Apologies for the confusion in the previous response.
You are correct. When t = 0, the equation y = -16t^2 + 40t - 5 simplifies to y = -5. This means that Luca’s initial launching point is 5 feet below street level. Thank you for pointing out the error.