The problem requires finding the average velocity of an object moving vertically along a line, given its position function $s(t) = -16t^2 + 96t$ over the interval from $t = 1$ to $t = 4$. The average velocity $\overline{v}$ can be calculated using the formula:

$\overline{v} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

Plugging in the values from the position function, where $t_1 = 1$ and $t_2 = 4$, we calculate:

1. Calculate $s(1)$: $s(1) = -16(1)^2 + 96(1) = -16 + 96 = 80 \, \text{meters}$

2. Calculate $s(4)$: $s(4) = -16(4)^2 + 96(4) = -16 \times 16 + 384 = -256 + 384 = 128 \, \text{meters}$

Now, find the average velocity: $\overline{v} = \frac{128 - 80}{4 - 1} = \frac{48}{3} = 16 \, \text{m/s}$

Therefore, the average velocity of the object over the interval from $t = 1$ to $t = 4$ is $16 \, \text{m/s}$, which corresponds to answer choice $[C] 16 \, \text{m/s}$.

Would you like further details on this problem, or do you have any other questions? Here are some related questions to consider:

1. What is the instantaneous velocity of the object at $t = 1$?
2. What is the instantaneous velocity of the object at $t = 4$?
3. How would the average velocity change if the interval was from $t = 0$ to $t = 4$?
4. Can you calculate the maximum height reached by the object?
5. What is the acceleration of the object at any time $t$?

Tip: When solving for average velocity over a time interval, always use the position values at the endpoints of the interval and divide by the difference in time to get the average rate of change.