This problem involves kinematics under constant acceleration, where the acceleration is due to gravity, $a=g$. The initial and final velocities are given as $v_A = 5$ m/s and $v_B = 15$ m/s, respectively.

[Q1] Find the distance between points A and B. We use the time-independent kinematic equation relating velocity, acceleration, and displacement.

$$v_f^2 = v_i^2 + 2a\Delta x$$

For this problem, $v_i = v_A$, $v_f = v_B$, $a = g$, and the distance is $\Delta x = d$.

$$v_B^2 = v_A^2 + 2gd$$

Solving for the distance $d$:

$$d = \frac{v_B^2 - v_A^2}{2g}$$

Substituting the given values:

$$d = \frac{(15 \text{ m/s})^2 - (5 \text{ m/s})^2}{2(10 \text{ m/s}^2)} = \frac{225 - 25}{20} \text{ m} = 10 \text{ m}$$

[Q2] Find the time taken to travel from A to B. We use the kinematic equation relating velocity, acceleration, and time.

$$v_f = v_i + at$$

Substituting the problem's variables ($v_i = v_A$, $v_f = v_B$, $a = g$):

$$v_B = v_A + gt$$

Solving for the time $t$:

$$t = \frac{v_B - v_A}{g}$$

Substituting the given values:

$$t = \frac{15 \text{ m/s} - 5 \text{ m/s}}{10 \text{ m/s}^2} = \frac{10}{10} \text{ s} = 1 \text{ s}$$