Given,
The height of the building, h=321 m
The acceleration due to gravity, g=9.8 m/s²
As the shoe was dropped, the initial velocity of the shoe is u=0 m/s
From the equation of motion,

Where t is the time it takes for the shoe to reach the ground.
On substituting the known values,
![\begin{gathered} 321=0+(1)/(2)*9.8* t^2 \\ \Rightarrow t=\sqrt[]{(2*321)/(9.8)} \\ t=8.09\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/b8xiw8xna1yk7qvd7qasrwpz1e3how3avv.png)
Thus it takes 8.09 s for the shoe to reach the ground.