To calculate the average velocity of the arrow over the time interval $[t_1, t_2]$, we use the formula for average velocity:

$v_{\text{avg}} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}$

where $h(t)$ is the height function. The height function is given by:

$h(t) = 61t - 0.83t^2$

We will compute the average velocity for each of the intervals: $[6,7]$, $[6,6.5]$, $[6,6.1]$, $[6,6.01]$, and $[6,6.001]$.

Let's start calculating.

Interval $[6, 7]$:

1. Compute $h(6)$ and $h(7)$: $h(6) = 61(6) - 0.83(6)^2$ $h(7) = 61(7) - 0.83(7)^2$
2. Compute the average velocity: $v_{\text{avg}} = \frac{h(7) - h(6)}{7 - 6}$

I'll calculate this and the rest step by step.Here are the average velocities for the given time intervals:

1. $[6, 7]$: $50.2100 \, \text{m/s}$
2. $[6, 6.5]$: $50.6250 \, \text{m/s}$
3. $[6, 6.1]$: $50.9570 \, \text{m/s}$
4. $[6, 6.01]$: $51.0317 \, \text{m/s}$
5. $[6, 6.001]$: $51.0392 \, \text{m/s}$

Would you like further details on any of these calculations, or have any other questions?

Here are five related questions for deeper understanding:

1. What is the instantaneous velocity of the arrow at $t = 6$ seconds?
2. How does the average velocity change as the interval size decreases?
3. What is the maximum height the arrow will reach?
4. How would the height function differ if the arrow was shot on Earth instead of the moon?
5. Can you calculate the time when the arrow will hit the ground?

Tip: Average velocity over smaller intervals tends to approach the instantaneous velocity at a specific point.