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=−16t2+40t−5
𝑦
=
−
16
𝑡
2
+
40
𝑡
−
5
. How far above or below street level is Luca’s initial launching point?(1 point)
Responses

5 feet below
5 feet below

5 feet above
5 feet above

0 feet above or below
0 feet above or below

2.5 feet below

Answer 1

To analyze the given scenario, let's break down the equation you provided:

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

In this equation:
- \( y \) represents the height of the object in feet,
- \( t \) represents the time in seconds,
- The coefficient \(-16\) indicates the effect of gravity (in this case, \( -32/2 = -16\) is used since we are measuring in feet and seconds).

The initial height from which Luca launches the object can be determined by evaluating the function when \( t = 0 \).

Plugging \( t = 0 \) into the equation:

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

This means that at \( t = 0 \) (the moment of launch), the height \( y \) is \(-5\) feet. Therefore, Luca’s initial launching point is **5 feet below street level**.

Thus, the correct response is:

**5 feet below**.