Avery strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.1 m/s. how long does he have to get out of the way if the shot was released at a height of 2.27 m, and he is 1.96 m tall?

We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)

Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)

-0.31=11.1t+(1/2)(-9.81)t^2
Rearrange and solve for t
-4.905t^2+11.1t-0.31=0
t=-0.02756 or t=2.291 seconds 
Reject the negative root to give
t=2.29 seconds (to 3 significant figures)