The average velocity of an object over a given time interval $[t_1, t_2]$ can be calculated using the formula:

$\text{Average velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}$

where $y(t) = 8t - 1.86t^2$ is the height function for the object, and $t_1$ and $t_2$ are the endpoints of the time interval.

For the interval $[1, 2]$, we'll substitute the values of $t_1 = 1$ and $t_2 = 2$ into the height function and compute the average velocity.

1. Find $y(1)$: $y(1) = 8(1) - 1.86(1^2) = 8 - 1.86 = 6.14 \, \text{m}$

2. Find $y(2)$: $y(2) = 8(2) - 1.86(2^2) = 16 - 1.86(4) = 16 - 7.44 = 8.56 \, \text{m}$

3. Calculate the average velocity: $\text{Average velocity} = \frac{y(2) - y(1)}{2 - 1} = \frac{8.56 - 6.14}{1} = 2.42 \, \text{m/s}$

Thus, the average velocity over the interval $[1, 2]$ is 2.42 m/s.

Would you like to calculate for other intervals or need further clarification?

Here are 5 related questions to consider:

1. What is the average velocity over the interval $[0, 1]$?
2. Can you find the instantaneous velocity at $t = 1$?
3. How does the acceleration affect the rock's velocity over time?
4. What is the maximum height reached by the rock?
5. How long does it take for the rock to hit the ground?

Tip: The average velocity formula is essentially the slope of the secant line between two points on the graph of the height function.