To find the change in position of the ball, we need to compute the displacement, which is the integral of the velocity function $v(t)$ with respect to time over the interval from $t = 0$ to $t = 6.6$ seconds.

The velocity function is given as:

$v(t) = 20 + 3t^3$

The change in position (displacement) $\Delta x$ is the integral of $v(t)$ from $t = 0$ to $t = 6.6$:

$\Delta x = \int_0^{6.6} v(t) \, dt = \int_0^{6.6} (20 + 3t^3) \, dt$

Now, let's compute this integral:

$\int (20 + 3t^3) \, dt = 20t + \frac{3t^4}{4}$

Now, we'll evaluate this from $t = 0$ to $t = 6.6$.

$\Delta x = \left[ 20t + \frac{3t^4}{4} \right]_0^{6.6}$

Substituting the limits $t = 6.6$ and $t = 0$:

$\Delta x = \left( 20(6.6) + \frac{3(6.6)^4}{4} \right) - \left( 20(0) + \frac{3(0)^4}{4} \right)$

Let me calculate that for you.The change in the position of the ball between $t = 0$ and $t = 6.6$ seconds is approximately $1555.11$ meters.

Do you need further details or have any questions?

Here are 5 related questions for further exploration:

1. How would the displacement change if the velocity function were different?
2. What is the physical interpretation of the integral of a velocity function?
3. How would you find the maximum velocity during the interval $t = 0$ to $t = 6.6$?
4. Can you compute the acceleration of the ball from the given velocity function?
5. How would you determine the average velocity over the same interval?

Tip: The area under a velocity-time graph gives the displacement, and integrating the velocity function serves the same purpose.