To find the average velocity of the ball over the time period beginning at $t = 3$ seconds and lasting $0.002$ seconds, we can use the formula for average velocity:

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

Here, $t_1 = 3$ seconds and $t_2 = t_1 + 0.002 = 3.002$ seconds.

Step 1: Calculate $y(t_1)$

First, we find the height at $t_1 = 3$ seconds:

$y(3) = 44(3) - 16(3)^2 = 132 - 144 = -12 \text{ feet}$

Step 2: Calculate $y(t_2)$

Next, find the height at $t_2 = 3.002$ seconds:

$y(3.002) = 44(3.002) - 16(3.002)^2$

Calculating each term:

$44(3.002) = 132.088$

$(3.002)^2 = 9.012004$

$16(9.012004) = 144.192064$

So,

$y(3.002) = 132.088 - 144.192064 = -12.104064 \text{ feet}$

Step 3: Compute the Average Velocity

Now, we can find the average velocity:

$\text{Average Velocity} = \frac{-12.104064 - (-12)}{3.002 - 3} = \frac{-0.104064}{0.002} = -52.032 \text{ ft/s}$

Final Answer:

The average velocity over the time period from $t = 3$ seconds to $t = 3.002$ seconds is $-52.032$ feet per second.

Would you like further explanation on any part of this calculation?

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Relative Questions:

1. What does the negative sign in the average velocity indicate?
2. How does the small time interval affect the accuracy of the average velocity?
3. How would the average velocity change if the interval was longer?
4. Can we use the instantaneous velocity instead of average velocity for small intervals?
5. How would you compute the average acceleration over the same time period?

Tip: The average velocity over a small time interval can approximate the instantaneous velocity at the midpoint of that interval, especially for smooth functions like quadratics.