This is a basic kinematics problem so it works the same mathematically as say the "car problem".
In the car problem, you can figure out the horizontal distance something travels using time by the relationship:
X = Vi(t) + 1/2(A)t^2
There are 4 variables in this equation: distance, velocity, time, and acceleration. If you know three of them, you can solve for the missing one. So let's say the car was initially at rest. Your initial velocity would be Vi = 0. Let's say the car accelerates at a constant rate of 4 m/s. Then A=4 m/s. And let's say that you want to know how far the car travelled in say 20 seconds. Now you know T. Once you know three of the 4, you can plug these in and use Algebra to solve for the missing distance traveled, X.
Your problem is very similar to the car problem except instead of a car traveling horizontally you have a ball travelling in a straight line vertically. Without spoiling anything you are given the time, the initial velocity, and the acceleration in your problem as well. Think about it a bit and you should be able to figure out what these three values are. From there, you can use the equation and solve.
u = 0 -as it is dropped.
t = 3.3 secs.
height = s
a = acceleration due to gravity = 10 approx.
s = ut +1/2at^2 ( t^2 is t square i.e t to the power of 2)
s = 0 + 1/2 * 10 * 3.3*3.3
s = 5 * 10.9
Therefore the height of the cliff is 54.5 m.