Let $s_n$ be the displacement during the $n$-th time interval $T$, and $v_n$ be the instantaneous velocity at the beginning of that interval. For motion with constant acceleration $a$, the kinematics equations apply.

[Q1] Can it be determined that the ball's motion is uniformly accelerated?

For an object in uniformly accelerated motion, the displacement during the $n$-th interval is:

$$s_n = v_n T + \frac{1}{2}aT^2$$

The displacement during the subsequent interval, $(n+1)$, is:

$$s_{n+1} = v_{n+1} T + \frac{1}{2}aT^2$$

The velocity at the beginning of the $(n+1)$-th interval is related to the velocity at the beginning of the $n$-th interval by $v_{n+1} = v_n + aT$.

Substituting $v_{n+1}$ into the equation for $s_{n+1}$:

$$s_{n+1} = (v_n + aT)T + \frac{1}{2}aT^2 = v_n T + aT^2 + \frac{1}{2}aT^2$$ $$s_{n+1} = \left(v_n T + \frac{1}{2}aT^2\right) + aT^2$$

Recognizing that the term in parentheses is $s_n$, we get:

$$s_{n+1} = s_n + aT^2$$

The difference in displacement between consecutive intervals is therefore:

$$\Delta s = s_{n+1} - s_n = aT^2$$

For uniformly accelerated motion, the acceleration $a$ is constant. Since the time interval $T$ is also constant, the difference $\Delta s$ should be constant.

Looking at the provided data, the values for $\Delta s$ (1.05, 1.05, 1.05, 1.14, 1.10, 1.09, 1.04, 1.09, 1.14 cm) are all close to each other, fluctuating around a central value. This approximate constancy of $\Delta s$ confirms that the ball's motion is uniformly accelerated, with the small variations likely due to experimental measurement errors.

[Q2] If it can be determined to be uniformly accelerated motion, find the acceleration of free fall, g, using the average value of $\Delta s$.

From the relationship derived above, the acceleration $g$ can be found using:

$$g = \frac{\Delta s}{T^2}$$

To obtain a more accurate result from the experimental data, we use the average value of $\Delta s$, denoted as $\overline{\Delta s}$. First, we calculate the average of the nine given $\Delta s$ values:

$$\overline{\Delta s} = \frac{1}{9} \sum_{i=1}^{9} (\Delta s)_i = \frac{1.05+1.05+1.05+1.14+1.10+1.09+1.04+1.09+1.14}{9} \text{ cm}$$ $$\overline{\Delta s} = \frac{9.75}{9} \text{ cm} \approx 1.083 \text{ cm}$$

Now we substitute this average value and the given time interval $T = 1/30$ s into the equation for $g$:

$$g = \frac{\overline{\Delta s}}{T^2} = \frac{(9.75/9) \text{ cm}}{(1/30 \text{ s})^2}$$ $$g = \frac{9.75}{9} \times (30)^2 \frac{\text{cm}}{\text{s}^2} = \frac{9.75}{9} \times 900 \frac{\text{cm}}{\text{s}^2} = 9.75 \times 100 \frac{\text{cm}}{\text{s}^2}$$ $$g = 975 \text{ cm/s}^2$$

Converting to standard SI units (m/s²):

$$g = 9.75 \text{ m/s}^2$$