Question

Luca is in a culvert below street level. He launches an object at an upward velocity of 40 feet per second. Graph the situation using the formula y=−16t2+40t−5 . How far above or below street level is Luca’s initial launching point?

Answer 1

To analyze the situation, we can interpret the formula provided, which is a quadratic equation describing the height \( y \) of an object over time \( t \):

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

In this equation:

- The term \(-16t^2\) represents the effect of gravity (in feet per second squared) acting on the object, which causes it to decelerate over time.
- The term \(40t\) represents the initial upward velocity of the object, which is 40 feet per second.
- The constant term \(-5\) represents the initial height of the object at \( t = 0 \). Since it's negative, this indicates that the initial height is below the street level.

### Step 1: Determine the Initial Height

To find out how far below street level Luca's initial launching point is, we can substitute \( t = 0 \) into the equation:

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

This means that the initial height of the launch point is \(-5\) feet, which indicates that the launching point is 5 feet below street level.

### Step 2: Graph the Situation

You can graph the quadratic function by plotting key points.

1. **Vertex (maximum height)**: The time at which the maximum height occurs can be found using the vertex formula \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 40 \):

\[
t = -\frac{40}{2 \times -16} = \frac{40}{32} = 1.25 \text{ seconds}
\]

2. **Maximum height \( y \)** at \( t = 1.25 \):

\[
y(1.25) = -16(1.25)^2 + 40(1.25) - 5
\]
\[
= -16(1.5625) + 50 - 5
\]
\[
= -25 + 50 - 5 = 20 \text{ feet}
\]

3. **Some additional points**:
- For \( t = 0 \): \( y(0) = -5 \) feet
- For \( t = 1 \):

\[
y(1) = -16(1)^2 + 40(1) - 5 = -16 + 40 - 5 = 19 \text{ feet}
\]

- For \( t = 2 \):

\[
y(2) = -16(2)^2 + 40(2) - 5 = -64 + 80 - 5 = 11 \text{ feet}
\]

- For \( t = 3 \):

\[
y(3) = -16(3)^2 + 40(3) - 5 = -144 + 120 - 5 = -29 \text{ feet}
\]

### Summary Results

- The graph would show a parabola opening downwards with:
- An initial height of -5 feet (5 feet below street level).
- A maximum height of 20 feet at 1.25 seconds.
- Points showing the trajectory of the object as it rises and then falls back down.

### Conclusion

Luca's initial launching point is 5 feet below street level. The graph of the function can be plotted with the above points to visualize the motion of the object over time.