The given problem asks for the average velocity of an arrow shot upward on the moon, where its height $h(t)$ in meters after $t$ seconds is given by the function:

$h(t) = 67t - 0.83t^2$

Average velocity formula:

The average velocity $v_{\text{avg}}$ over a time interval $[t_1, t_2]$ is calculated using the formula:

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

Time intervals:

The time intervals given are:

• $[5, 6]$
• $[5, 5.5]$
• $[5, 5.1]$
• $[5, 5.01]$
• $[5, 5.001]$

We will compute the average velocity for each of these intervals using the provided function for height.

Let me calculate these values.Here are the average velocities for each time interval, calculated to four decimal places:

1. For $[5, 6]$: $57.8700 \, \text{m/s}$
2. For $[5, 5.5]$: $58.2850 \, \text{m/s}$
3. For $[5, 5.1]$: $58.6170 \, \text{m/s}$
4. For $[5, 5.01]$: $58.6917 \, \text{m/s}$
5. For $[5, 5.001]$: $58.6992 \, \text{m/s}$

Let me know if you would like further details or clarifications!

Here are five related questions to consider:

1. What is the physical interpretation of average velocity in this context?
2. How would the velocity change as the time intervals become smaller?
3. Can you derive the instantaneous velocity from the height function?
4. How does gravity on the moon (0.83 $m/s^2$) affect the arrow’s motion?
5. How would the equation change if this was happening on Earth?

Tip: The smaller the time interval, the closer the average velocity approximates the instantaneous velocity.