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=−16t^2+40t−5
. How far above or below street level is Luca’s initial launching point?(1 point)
Responses

2.5 feet below
2.5 feet below

5 feet below
5 feet below

0 feet above or below
0 feet above or below

5 feet above

Answer 1

To determine how far above or below street level Luca's initial launching point is, we need to analyze the quadratic equation provided:

\[ y = -16t^2 + 40t - 5 \]

In this equation:
- \( y \) represents the height of the object above (positive) or below (negative) street level.
- \( t \) represents time in seconds.
- The constant term, in this case \(-5\), indicates the initial height of the object when \( t = 0 \).

To find the initial launching point (when \( t = 0 \)), we can substitute \( t = 0 \) into the equation:

\[ y(0) = -16(0)^2 + 40(0) - 5 \]
\[ y(0) = -5 \]

Thus, the initial height, or the position of the launching point, is 5 feet below street level.

So the correct response is:

**5 feet below**.