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,
[tex]h=(u\times t)+\frac{1}{2}gt^2[/tex]
Where t is the time it takes for the shoe to reach the ground.
On substituting the known values,
[tex]\begin{gathered} 321=0+\frac{1}{2}\times9.8\times t^2 \\ \Rightarrow t=\sqrt[]{\frac{2\times321}{9.8}} \\ t=8.09\text{ s} \end{gathered}[/tex]
Thus it takes 8.09 s for the shoe to reach the ground.
