Let $H$ be the total height of the tower and $t_f$ be the total time of fall. The stone is dropped from rest, so its initial velocity is $v_0 = 0$. We use the standard kinematic equation for position under constant acceleration $g$, with the origin at the top of the tower and the downward direction as positive.

The total height of the tower is the distance fallen in time $t_f$:

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

The distance fallen in the first $(t_f - 1)$ seconds is:

$$H_{t_f-1} = \frac{1}{2}g(t_f - 1)^2$$

The distance traveled in the final second, $d_{last} = 45 \text{ m}$, is the difference between the total height and the height fallen up to one second before impact:

$$d_{last} = H - H_{t_f-1}$$ $$d_{last} = \frac{1}{2}gt_f^2 - \frac{1}{2}g(t_f - 1)^2$$

Now, we derive an expression for the total time of fall, $t_f$.

$$d_{last} = \frac{1}{2}g [t_f^2 - (t_f^2 - 2t_f + 1)]$$ $$d_{last} = \frac{1}{2}g (2t_f - 1)$$ $$2d_{last} = g(2t_f - 1)$$ $$2t_f - 1 = \frac{2d_{last}}{g}$$ $$t_f = \frac{d_{last}}{g} + \frac{1}{2}$$

Substituting the given values:

$$t_f = \frac{45 \text{ m}}{10 \text{ m/s}^2} + \frac{1}{2} \text{ s} = 4.5 \text{ s} + 0.5 \text{ s} = 5 \text{ s}$$

[Q1] With the total time of fall, we can find the total height of the tower.

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

Substituting values:

$$H = \frac{1}{2}(10 \text{ m/s}^2)(5 \text{ s})^2 = 5 \cdot 25 \text{ m} = 125 \text{ m}$$

[Q2] The velocity of the stone just before it hits the ground, $v_f$, is given by:

$$v_f = v_0 + gt_f = gt_f$$

Substituting values:

$$v_f = (10 \text{ m/s}^2)(5 \text{ s}) = 50 \text{ m/s}$$