To find the average velocity of the arrow over the specified time intervals, we use the formula for average velocity:

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

where $t_1$ is the initial time and $t_2$ is the final time. Here, $h(t) = 44t - 0.83t^2$ is the height function.

We will now calculate the average velocity over each of the given intervals.

Interval [8, 9]:

• $t_1 = 8$, $t_2 = 9$

$\text{Average velocity} = \frac{h(9) - h(8)}{9 - 8}$

First, calculate $h(9)$ and $h(8)$:

$h(9) = 44(9) - 0.83(9)^2 = 396 - 67.23 = 328.77$ $h(8) = 44(8) - 0.83(8)^2 = 352 - 53.12 = 298.88$

$\text{Average velocity} = \frac{328.77 - 298.88}{1} = 29.89 \, \text{m/s}$

Interval [8, 8.5]:

• $t_1 = 8$, $t_2 = 8.5$

$\text{Average velocity} = \frac{h(8.5) - h(8)}{8.5 - 8}$

Calculate $h(8.5)$:

$h(8.5) = 44(8.5) - 0.83(8.5)^2 = 374 - 60.07625 = 313.92375$

$\text{Average velocity} = \frac{313.92375 - 298.88}{0.5} = \frac{15.04375}{0.5} = 30.0875 \, \text{m/s}$

Interval [8, 8.1]:

• $t_1 = 8$, $t_2 = 8.1$

$\text{Average velocity} = \frac{h(8.1) - h(8)}{8.1 - 8}$

Calculate $h(8.1)$:

$h(8.1) = 44(8.1) - 0.83(8.1)^2 = 356.4 - 54.47283 = 301.92717$

$\text{Average velocity} = \frac{301.92717 - 298.88}{0.1} = \frac{3.04717}{0.1} = 30.4717 \, \text{m/s}$

Interval [8, 8.01]:

• $t_1 = 8$, $t_2 = 8.01$

$\text{Average velocity} = \frac{h(8.01) - h(8)}{8.01 - 8}$

Calculate $h(8.01)$:

$h(8.01) = 44(8.01) - 0.83(8.01)^2 = 352.44 - 53.1846603 = 299.2553397$

$\text{Average velocity} = \frac{299.2553397 - 298.88}{0.01} = \frac{0.3753397}{0.01} = 37.53397 \, \text{m/s}$

Interval [8, 8.001]:

• $t_1 = 8$, $t_2 = 8.001$

$\text{Average velocity} = \frac{h(8.001) - h(8)}{8.001 - 8}$

Calculate $h(8.001)$:

$h(8.001) = 44(8.001) - 0.83(8.001)^2 = 352.044 - 53.1278477 = 298.9161523$

$\text{Average velocity} = \frac{298.9161523 - 298.88}{0.001} = \frac{0.0361523}{0.001} = 36.1523 \, \text{m/s}$

Summary of Average Velocities:

1. Interval [8, 9]: $29.89 \, \text{m/s}$
2. Interval [8, 8.5]: $30.0875 \, \text{m/s}$
3. Interval [8, 8.1]: $30.4717 \, \text{m/s}$
4. Interval [8, 8.01]: $37.53397 \, \text{m/s}$
5. Interval [8, 8.001]: $36.1523 \, \text{m/s}$

Would you like more details on any of the steps?

Related questions:

1. How do we calculate instantaneous velocity using limits?
2. What is the significance of smaller time intervals in average velocity?
3. How does the acceleration on the moon affect the motion of the arrow?
4. Can we use a different method to find the average velocity?
5. What happens to velocity as the arrow approaches its peak height?

Tip:

The closer the time intervals are, the more the average velocity approximates the instantaneous velocity.