Let the positive direction be downwards, with the origin at the roof. The height of the building $H$ is the total distance the ball falls. We can find this from the final velocity $v_g$ just before impact, using $H = v_g^2 / (2g)$ since the ball starts from rest.

First, find the velocity $v_1$ at the top of the window. For the motion across the window:

$$h_w = v_1 t_w + \frac{1}{2}gt_w^2 \implies v_1 = \frac{h_w}{t_w} - \frac{1}{2}gt_w$$

Next, find the velocity $v_2$ at the bottom of the window:

$$v_2 = v_1 + gt_w = \left(\frac{h_w}{t_w} - \frac{1}{2}gt_w\right) + gt_w = \frac{h_w}{t_w} + \frac{1}{2}gt_w$$

The time to fall from the bottom of the window to the sidewalk is $t_{fall} = T/2$, due to the symmetric nature of the bounce. Now, find the final velocity $v_g$ at the sidewalk:

$$v_g = v_2 + g t_{fall} = \left(\frac{h_w}{t_w} + \frac{1}{2}gt_w\right) + g\frac{T}{2} = \frac{h_w}{t_w} + \frac{g}{2}(t_w + T)$$

Finally, the total height of the building $H$ is related to the final velocity by the kinematic equation $v_g^2 = v_0^2 + 2gH$. Since $v_0 = 0$:

$$H = \frac{v_g^2}{2g} = \frac{1}{2g}\left[ \frac{h_w}{t_w} + \frac{g}{2}(t_w + T) \right]^2$$