As the ball is "thrown vertically down from the edge" of the cliff, the distance fallen will be the height of the cliff.

Let's use the formula #s = u t + frac(1)(2) a t^(2)#; where #s# is the distance, #u# is the initial velocity, #a# is the acceleration, and #t# is the time taken.

In this case, #a# will be equal to #9.81# #"m/s"^(2)#, which is the acceleration due to gravity:

#Rightarrow s = 0# #"m/s"# #times 6# #"s"# #+ frac(1)(2) times 9.81# #"m/s"^(2) times 6^(2)# #"s"^(2)#

#Rightarrow s = 4.905# #"m/s"^(2) times 36# #"s"^(2)#

#therefore s = 176.58# #"m"#

Therefore, the cliff is around #177# #"m"# high.