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

5 feet above
5 feet above

5 feet below
5 feet below

0 feet above or below
0 feet above or below

2.5 feet below
2.5 feet below

1 answer

To determine how far above or below street level Luca’s initial launching point is, we can look at the function provided:

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

In this equation:

  • \( y \) represents the height of the object above or below street level,
  • \( t \) represents time in seconds,
  • The term \( -5 \) indicates the initial height from which the object is launched at \( t = 0 \).

Now, to find the height at the time of launch (when \( t = 0 \)), we can substitute \( t = 0 \) into the equation:

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

This means that the initial height of Luca’s launching point is -5 feet, indicating that he is 5 feet below street level.

Therefore, the answer is:

5 feet below.