### Steel Ball

Write the speed of the ball at the top of the window as $v_{1}$ and at the bottom of the window $v_{2}.$
Call the time between when the ball is at the top of the window and at the bottom of the window $t_{12}.$
So we know $t_{12}=0.125s.$
These speeds are related: $v_{2}=v_{1}-gt_{12}.$ The average speed is $v_{av}=\frac{v_{1}+v_{2}}{2}=v_{1}-(\frac{1}{2}gt_{12}).$
The distance traveled in the time $t_{12}$ is $y_{12}=-120m.$ $y_{12}=v_{av}t_{12}-\frac{1}{2}gt_{12}^{2}.$
We can solve this for $v_{1}$ $v_{1} = \frac{y_{12}+(1/2)gt_{12}^{2}}{t_{12}} = \frac{(-1.20m)+(9.81\frac{m}{s^{2}})(0.125s)^{2}/2}{0.125s}=-8.99\frac{m}{s}.$
Now that we know $v_{1,}$ we can find the height of the building above the top of the window. The time that the object has fallen to get to the top of the window is $t_{1}=\frac{-v_{1}}{g}=-(-8.99\frac{m}{s})(9.81\frac{m}{s^{2}})=0.916m.$
The total time for falling is then (0.916s)+(0.125s)+(1.0s)=2.04s. Note that we had to divide the last time by two!
The total distance from the top of the building to the bottom is then $y=-\frac{1}{2}gt^{2}=-\frac{1}{2}(9.81\frac{m}{s^{2}})(2.04s)^{2}=20.4m.$