We define the downward direction as positive. The acceleration due to gravity is $a = g$.

[Q1] Find the distance between points A and B. We use the time-independent kinematic equation relating final velocity ($v_B$), initial velocity ($v_A$), acceleration ($g$), and displacement ($\Delta y_{AB}$).

$$v_B^2 = v_A^2 + 2g\Delta y_{AB}$$

Solving for the distance $\Delta y_{AB}$:

$$\Delta y_{AB} = \frac{v_B^2 - v_A^2}{2g}$$

Substituting the given values:

$$\Delta y_{AB} = \frac{(25 \text{ m/s})^2 - (20 \text{ m/s})^2}{2(10 \text{ m/s}^2)} = \frac{625 - 400}{20} \text{ m} = \frac{225}{20} \text{ m}$$ $$\Delta y_{AB} = 11.25 \text{ m}$$

[Q2] Find the initial height from which the ball was dropped. First, we find the velocity of the ball when it reaches the ground, $v_G$. We consider the motion from point B to the ground (G), which takes time $t_{BG} = 0.5$ s.

$$v_G = v_B + gt_{BG}$$

Next, we find the total initial height, $H$. Since the ball was dropped, its initial velocity was $v_0 = 0$. We relate the final velocity $v_G$ to the total height $H$ using the time-independent kinematic equation over the entire fall.

$$v_G^2 = v_0^2 + 2gH$$

With $v_0=0$, we have $v_G^2 = 2gH$. Substituting the expression for $v_G$:

$$(v_B + gt_{BG})^2 = 2gH$$

Solving for the total height $H$:

$$H = \frac{(v_B + gt_{BG})^2}{2g}$$

Substituting the given values:

$$H = \frac{(25 \text{ m/s} + (10 \text{ m/s}^2)(0.5 \text{ s}))^2}{2(10 \text{ m/s}^2)} = \frac{(25 + 5)^2}{20} \text{ m} = \frac{(30)^2}{20} \text{ m}$$ $$H = \frac{900}{20} \text{ m} = 45 \text{ m}$$