We analyze the motion of the ball using the equations for uniformly accelerated motion. We define the origin at the release point, with the downward direction as positive. The initial velocity is $v_0 = 0$ and the acceleration is $a = g = 9.8 \text{ m/s}^2$. The total height of the fall is $H = 19.6 \text{ m}$.

[Q1] How long does it take for the ball to reach the ground?

The vertical position $y$ at time $t$ is given by the kinematic equation:

$$y(t) = v_0 t + \frac{1}{2}gt^2$$

For the entire fall, the displacement is $H$ and the time is the total time $T$. With $v_0 = 0$, we have:

$$H = \frac{1}{2}gT^2$$

We derive the expression for the total time of fall, $T$:

$$T = \sqrt{\frac{2H}{g}}$$

Substituting the given values:

$$T = \sqrt{\frac{2(19.6 \text{ m})}{9.8 \text{ m/s}^2}} = \sqrt{4.0 \text{ s}^2} = 2.0 \text{ s}$$

[Q2] What is the ball's displacement during the last 1 s of its fall?

The displacement in the last second, $\Delta y_{last}$, is the total displacement $H$ minus the displacement during the first $(T - 1 \text{ s})$ of the fall.

$$\Delta y_{last} = y(T) - y(T - 1 \text{ s})$$

Using the position function $y(t) = \frac{1}{2}gt^2$:

$$\Delta y_{last} = \frac{1}{2}gT^2 - \frac{1}{2}g(T - 1 \text{ s})^2$$

Let $t_L = 1 \text{ s}$ be the duration of the final interval. We factor the expression:

$$\Delta y_{last} = \frac{1}{2}g[T^2 - (T - t_L)^2]$$

Expanding the squared term:

$$\Delta y_{last} = \frac{1}{2}g[T^2 - (T^2 - 2Tt_L + t_L^2)] = \frac{1}{2}g(2Tt_L - t_L^2)$$

This simplifies to:

$$\Delta y_{last} = gt_L(T - \frac{t_L}{2})$$

Substituting $t_L = 1 \text{ s}$, $T = 2.0 \text{ s}$, and $g = 9.8 \text{ m/s}^2$:

$$\Delta y_{last} = (9.8 \text{ m/s}^2)(1 \text{ s})(2.0 \text{ s} - \frac{1 \text{ s}}{2})$$ $$\Delta y_{last} = (9.8 \text{ m/s})(1.5 \text{ s}) = 14.7 \text{ m}$$