Let $H$ be the total height of the tower and $t$ be the total time of fall. The object is dropped from rest, so its initial velocity is $v_0 = 0$. The acceleration is $a=g$.

The distance an object falls from rest in time $\tau$ is given by the kinematic equation:

$$d(\tau) = \frac{1}{2}g\tau^2$$

[Q1] Find the height of the tower. The total height $H$ is the distance fallen in the total time $t$:

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

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

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

The distance traveled during the final second of fall, $\Delta H_{last}$, is the difference between the total height and the distance fallen up to the second-to-last second:

$$\Delta H_{last} = H - H_{t-1} = \frac{1}{2}gt^2 - \frac{1}{2}g(t-1)^2$$

Expanding and simplifying the expression:

$$\Delta H_{last} = \frac{1}{2}g[t^2 - (t^2 - 2t + 1)] = \frac{1}{2}g(2t-1)$$

The problem states that this distance is 9/25 of the total height:

$$\Delta H_{last} = \frac{9}{25}H$$

Substitute the derived expressions for $\Delta H_{last}$ and $H$:

$$\frac{1}{2}g(2t-1) = \frac{9}{25}\left(\frac{1}{2}gt^2\right)$$

The term $\frac{1}{2}g$ cancels out, leaving a quadratic equation for $t$:

$$2t-1 = \frac{9}{25}t^2$$ $$9t^2 - 50t + 25 = 0$$

Factoring the quadratic equation:

$$(9t-5)(t-5) = 0$$

This gives two possible solutions for $t$: $t = 5/9$ s and $t = 5$ s. For the concept of a "final second" to be physically meaningful, the total time of fall must be greater than 1 second. Therefore, we choose the valid solution:

$$t = 5 \text{ s}$$

Now, substitute this time back into the equation for the total height $H$:

$$H = \frac{1}{2}gt^2 = \frac{1}{2}g(5)^2 = \frac{25}{2}g$$

Using the standard value for the acceleration due to gravity, $g = 9.8$ m/s²:

$$H = \frac{25}{2}(9.8 \text{ m/s}^2) = 122.5 \text{ m}$$