The flowerpot is in free fall, so it moves by uniformly accelerated motion with acceleration equal to [tex]g=9.81 m/s^2[/tex]. Its initial speed is zero, so we can write the law of motion for the pot as
[tex]S(t) = \frac{1}{2}gt^2 [/tex]
where S(t) is the distance covered by the pot after a time t.
We want to find the time t at which the pot reaches ground, so the time t after which the pot has covered 9.2 m, therefore we should require S(t)=9.2 m and find t:
[tex]9.2 m= \frac{1}{2} gt^2[/tex]
[tex]t= \sqrt{ \frac{2 \cdot 9.2m}{9.81 m/s^2} }=1.37 s [/tex]