It takes 1.8s to drop a ball of a bridge. How high is the bridge and what speed is reached?

t= 1.8s 

a=-9.81ms^-2

v=u+at 

v= -9.81 x 1.8 =17.658 m/s

s= ut + 1/2 at²  (as u is 0 s= 1/2 at²)

s = 1/2 x 9.81 x 1.8² = 15.9m