A ball is dropped from rest from the top of a tall cliff. In the...

Let the velocity of the ball $1\text{ s}$ before hitting the ground be $u$.

For the last $1\text{ s}$ of the fall, the distance covered is $45\text{ m}$. Using the second equation of motion:
$s = ut + \dfrac{1}{2}gt^2$
$45 = u(1) + \dfrac{1}{2}(10)(1)^2$
$45 = u + 5 \Rightarrow u = 40\text{ m s}^{-1}$

This velocity $u = 40\text{ m s}^{-1}$ is attained by the ball after falling from rest. Let the time taken to reach this velocity be $t_1$.
Using $v = u_{initial} + gt_1$:
$40 = 0 + 10t_1 \Rightarrow t_1 = 4\text{ s}$

The total time of fall is the time taken to reach $40\text{ m s}^{-1}$ plus the last $1\text{ s}$:
$t_{total} = 4 + 1 = 5\text{ s}$

The total height of the cliff $H$ is the distance fallen from rest in $5\text{ s}$:
$H = \dfrac{1}{2}gt_{total}^2$
$H = \dfrac{1}{2}(10)(5)^2 = 5 \times 25 = 125\text{ m}$

Answer: $125\text{ m}$