The distance traveled by an object that starts with an initial velocity u and travels for a time t under an acceleration a is given by s = u*t + (1/2)*a*t^2. If the same formula is used for an object dropped from an high building, the distance traveled s is the height of the building, the initial velocity is 0 and the acceleration a = 9.8 m/s^2. This gives the height of the building h = 0 + (1/2)*9.8*t^2
The ball dropped from the roof of the building takes 4 s to reach the street. The height of the building is h = 0*4 + (1/2)*9.8*4^2 = 4.9*16 = 78.4 m
The height of a building from the roof of which if a ball is dropped it takes 4 s to reach the base is 78.4 m.