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Let's analyze the problem step by step.

### Given Information:
- A projectile is launched from ground level with an initial velocity \( v_0 \) feet per second.
- Neglecting air resistance, its height \( y \) in feet after \( t \) seconds is given by the equation:
  \[
  y = -16t^2 + v_0 t
  \]
  
### Questions to Answer:
1. **(a) Find the time(s) that the projectile will reach a height of 80 ft when \( v_0 = 96 \) feet per second.**
2. **(b) Determine the time(s) it will return to the ground (i.e., \( y = 0 \)) for \( v_0 = 96 \) feet per second.**

### Solution Approach:

#### Part (a): Finding the time(s) to reach a height of 80 ft
1. Set the height equation equal to 80 ft:
   \[
   -16t^2 + 96t = 80
   \]
2. Rearrange the equation:
   \[
   -16t^2 + 96t - 80 = 0
   \]
   Simplify by dividing the entire equation by -16:
   \[
   t^2 - 6t + 5 = 0
   \]
3. Factor the quadratic equation:
   \[
   (t - 5)(t - 1) = 0
   \]
4. Solve for \( t \):
   \[
   t = 5 \text{ seconds}, \quad t = 1 \text{ second}
   \]
   So, the projectile reaches 80 ft at \( t = 1 \) second and \( t = 5 \) seconds.

#### Part (b): Finding the time to return to the ground
1. Set the height equation equal to 0 (since it returns to the ground at \( y = 0 \)):
   \[
   -16t^2 + 96t = 0
   \]
2. Factor the equation:
   \[
   t(-16t + 96) = 0
   \]
3. Solve for \( t \):
   \[
   t = 0 \text{ seconds}, \quad t = 6 \text{ seconds}
   \]
   The projectile returns to the ground at \( t = 6 \) seconds (we disregard \( t = 0 \) because it's the initial launch time).

### Answers:
- **Part (a)**: The correct choice is **A**. The projectile will reach a height of 80 ft at **1 second and 5 seconds**.
- **Part (b)**: The projectile returns to the ground at **6 seconds**.

Would you like any further details or have any questions?

### Follow-Up Questions:
1. How does changing the initial velocity affect the time to reach a specific height?
2. What would be the maximum height reached by the projectile with \( v_0 = 96 \) feet per second?
3. How would the time to reach 80 feet change if the initial velocity were doubled?
4. Can the equation be used to determine when the projectile reaches a specific height other than 80 feet?
5. How does gravity factor into the height equation provided?

**Tip:** When solving quadratic equations, always check if the equation can be factored before using the quadratic formula. It can save time and simplify the solution process.

Explore projectile motion with an initial velocity of 96 feet per second. Solve for the time to reach a height of 80 feet and determine when the projectile returns to the ground. Detailed step-by-step solutions provided.

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