Let initial velocity be given by #'u'# = #6.0ms^-1#,

and

Final velocity be given by #'v' = 0 ms^-1#.

#'t' = 3.0s# is the time taken by the stone to reach the ground, and

#'a' = 9.81ms^-2 # is the acceleration due to gravity.

Let #'s'# be the distance covered.

According to the position- time relationship or the second equation of motion:

#s=ut +1/2at^2#

#s = 6.0 m/cancels xx 3.0cancels + 1/2 xx 9.81m/cancels^2 xx (3.0)^2 cancels^2#

#s = 18 m + (9.81 xx 9)/2 m#

#s= 18m + 44.145m#

#s= 62.145 m# is the distance covered by the stone from top of the cliff to the bottom.

#therefore# The height of the cliff is #62.145 m#