$$s=ut+\frac{1}{2}at^2$$
Where,
s is the distance fallen (in meters)
u is the initial velocity (in meters per second)
a is the acceleration due to gravity (in meters per second squared)
t is the time taken (in seconds)
In this case, the object is dropped from rest, so its initial velocity is 0 m/s. The acceleration due to gravity is 9.8 m/s^2. And the time taken for the object to fall 128 m can be found using the formula:
$$s=ut+\frac{1}{2}at^2$$
$$128=0+\frac{1}{2}(9.8)t^2$$
$$t^2=\frac{128}{4.9}$$
$$t^2=26$$
$$t= \sqrt{26} = 5.1 \ s$$
Now, the distance fallen during the final second can be found by substituting t = 5 s and t = 4 s into the equation of motion:
$$s=ut+\frac{1}{2}at^2$$
$$s=0(5)+\frac{1}{2}(9.8)(5^2)$$
$$s= \frac{1}{2}(9.8)(25) = 122.5 \ m$$
Therefore, the distance fallen during its final second in air is 122.5 m.
