This question uses the equations of motion, commonly referred to as the SUVAT equations. We will need to consider the horizontal and vertical components of motion separately. As the vertical component determines the time in the air and is also affected by gravitational acceleration we will consider this first as this determines the time in the air for the cannonball.
For this question I have taken upwards as positive and the top of the cliff as zero meaning the ground is at a displacement/distance of -50m
The vertical component of acceleration can be found using trigonometry by taking the velocity vector and creating a right angled triangle. This gives us a inital vertical velocity uy = 20sin(60) ms-1 , we know gravity acts downwards and will act on the cannonball giving us an acceleration ay = -9.8 ms-2 we need to travel a distance s = -50 m (notice the sign as we have taken upwards as positive)
The equation of motion needed is s = ut + 0.5at2. Subbing in the numbers: - 50 = 20sin(60) - 4.9t2 This is a quadratic and will have 2 solutions for the time which is what we need to calculate, however as this is considering a parabola one of these will be a negative value and should be ignored.
Using the quadratic formula we get the non negative value of t = 5.418122.. (always keep full calculator answers until the very end to avoid rounding errors, I like to store them in the memory)
Now we have the amount of time the cannonball is in the air, we can calculate how far this corresponds to horizontally. We will neglect air resistance.
Using the same equation of motion as above, but with no acceleration this simplifies the equation to s = u*t
using the triangle above for the horizontal velocity we get ux = 20cos(60) and subbing in our value for t (5.4181...) we get a horizontal distance of sx = 54.181... m
Rounding this to 2 decimal places we get our final answer:
s = 54.18 m