To solve this, we need to compute the average velocity over the given time intervals. The average velocity formula is:

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

Where:

• $h(t) = 41t - 0.83t^2$ is the height function
• $t_1$ and $t_2$ represent the time points at the start and end of each interval.

We will compute the average velocity for the following time intervals:

1. $[9, 10]$
2. $[9, 9.5]$
3. $[9, 9.1]$
4. $[9, 9.01]$
5. $[9, 9.001]$

Let's proceed with these calculations. Given function for height def h(t): return 41 * t - 0.83 * t**2

Function to calculate average velocity for the time interval [t1, t2]

def average_velocity(t1, t2): return (h(t2) - h(t1)) / (t2 - t1)

Time intervals

time_intervals = [(9, 10), (9, 9.5), (9, 9.1), (9, 9.01), (9, 9.001)]

Calculate average velocities for each time interval

avg_velocities = [average_velocity(t1, t2) for t1, t2 in time_intervals]

avg_velocitiesHere are the average velocities for each of the given time intervals:

1. $[9, 10]$: 25.230 m/s
2. $[9, 9.5]$: 25.645 m/s
3. $[9, 9.1]$: 25.977 m/s
4. $[9, 9.01]$: 26.052 m/s
5. $[9, 9.001]$: 26.059 m/s

These results are rounded to three decimal places as requested.

Would you like to go over any of the calculations in more detail, or need help with anything else?

Here are some related questions you might find interesting:

1. How can we interpret the average velocity at smaller time intervals?
2. What is the instantaneous velocity at $t = 9$?
3. How does the acceleration of the arrow change over time?
4. What is the maximum height of the arrow?
5. How can you derive the velocity function from the given height function?

Tip: When calculating average velocity over small intervals, it gives an approximation of instantaneous velocity, especially as the time intervals get smaller!