The problem you posted involves solving for the time at which a ball strikes the ground and when it passes the top of the building. The equation for the ball's height at any time $t$ is given by:

$s(t) = 128 + 32t - 16t^2$

(a) When does the ball strike the ground?

The ball strikes the ground when $s(t) = 0$. So, we need to solve:

$128 + 32t - 16t^2 = 0$

(b) When does the ball pass the top of the building on the way down?

The ball is thrown from the top of the building at $s = 128$. We need to find when the ball's height again equals 128 feet on its way down. Thus, we solve:

$128 + 32t - 16t^2 = 128$

I will now solve these equations for you.### (a) Time when the ball strikes the ground: The solutions for when the ball strikes the ground are $t = -2$ and $t = 4$. Since time cannot be negative, the ball strikes the ground at $t = 4$ seconds.

(b) Time when the ball passes the top of the building on the way down:

The solutions for when the ball is at the height of 128 feet (the top of the building) are $t = 0$ and $t = 2$. The ball is initially at the top at $t = 0$, and it passes the top again on its way down at $t = 2$ seconds.

Let me know if you'd like further details on any part of the solution!

Here are 5 related questions for further exploration:

1. What would happen to the time if the initial velocity were doubled?
2. How would the result change if the building height were different?
3. Can we derive a general formula for time to hit the ground for any initial velocity?
4. What is the velocity of the ball just before it hits the ground?
5. How would air resistance affect the times?

Tip: When solving quadratic equations, always discard negative time values in real-world problems.