The image presents a physics problem concerning an object in free fall with a function $s(t) = -16t^2 + 96t$ representing the height of the object over time.

To summarize the problem:

1. A ball is dropped from rest at $t = 0$, with the position function $s(t) = -16t^2 + 96t$, where $s(t)$ is the height of the ball at time $t$.
2. The task is to calculate the average velocity over the interval from $t = 3$ to $t = 5$.

The key steps are outlined as:

• The height at $t = 3$ and $t = 5$ is calculated using the function $s(t)$. $s(3) = -16(3^2) + 96(3) = 144 \text{ feet}$ $s(5) = -16(5^2) + 96(5) = 80 \text{ feet}$
• The average velocity is calculated using the formula: $\text{Average velocity} = \frac{s(5) - s(3)}{5 - 3}$ Substituting the values: $\text{Average velocity} = \frac{80 - 144}{5 - 3} = \frac{-64}{2} = -32 \, \text{feet per second}.$

Would you like further details on this process or have any additional questions?

Here are some related questions to expand your understanding:

1. How do you calculate instantaneous velocity from a position function?
2. What is the significance of the negative sign in the average velocity result?
3. How would you compute the velocity of the ball at a specific time using calculus?
4. What physical conditions must be true for the ball's height to reach zero?
5. How would the average velocity change if the time interval were shortened to one second?

Tip: The derivative of a position function $s(t)$ gives the instantaneous velocity function $v(t)$, which can be used for deeper analysis of motion.