Answer
The height of the building is $~~20.5~m$
Work Step by Step
Let $v_t$ be the speed at the top of the window.
Let $v_b$ be the speed at the bottom of the window.
Then:
$v_b = v_t+at$
$v_b = v_t+(9.8)(0.125)$
$v_b = v_t+1.225$
Let $v_{ave}$ be the average speed while passing the window.
We can find $v_t$:
$v_{ave}~t = \Delta y$
$\frac{v_t+v_b}{2} = \frac{\Delta y}{t}$
$\frac{v_t+v_t+1.225}{2} = \frac{\Delta y}{t}$
$v_t+0.6125 = \frac{\Delta y}{t}$
$v_t = \frac{\Delta y}{t}-0.6125$
$v_t = \frac{1.20}{0.125}-0.6125$
$v_t = 9.0~m/s$
We can find $y_1$, the distance from the roof to the top of the window:
$v_t^2 = v_0+2ay_1$
$v_t^2 = 0+2ay_1$
$y_1 = \frac{v_t^2}{2a}$
$y_1 = \frac{(9.0~m/s)^2}{(2)(9.8~m/s^2)}$
$y_1 = 4.13~m$
We can find $v_b$:
$v_b = v_t+1.225$
$v_b = 9.0+1.225$
$v_b = 10.225~m/s$
Note that it takes $1.00~s$ to fall from the bottom of the window to the ground.
We can find $y_2$, the distance from the bottom of the window to the ground:
$y_2 = v_b~t+\frac{1}{2}at^2$
$y_2 = (10.225~m/s)(1.00~s)+\frac{1}{2}(9.8~m/s^2)(1.00~s)^2$
$y_2 = 15.125~m$
We can find the height of the building:
$y = 4.13~m+1.20~m+15.125~m$
$y = 20.5~m$
The height of the building is $~~20.5~m$
