Physics - Motion question

A hammer mass 30 kg, was tied to a 5.0m long rope. It swings from rest at position X and falls to position Y...
at the center was the other end of the rope, and to the right is the end tied to the hammer, while below the center is Y, there center, as A, AX = AY and they are perpendicular... please tell me if you don't get the question...
Given gravitational force = 10 m s^-2

what is the speed of the hammer on reaching position Y?
(is the displacement is something to do with circumference or what?)

Ok: AX = AY means the hammer starts from horizontal to the ground. Ie. it swings through 9 degrees before it gets to the bottom. Since the rope is 5.0m long, this means it falls a distance of 5.0m vertically.
This is a conservation of energy question.
So, (1/2)mv^2 = mgh. Ie. initial gravitational potential energy = final kinetic energy.
If we keep working algebratically and solve for v, we get:
2mgh = mv^2
2gh = v^2
sqrt(2gh) = v
Therefore, v = sqrt(2*10*5) = sqrt(100) = 10m/s!
The other way to approach this is to find the initial gravitational potential energy, and then make kinetic energy equal to this and solve.
Ie. mgh = 30*10*5 = 1500 J or energy at top of swing.
Therefore, 1500 = (1/2)mv^2
(1500*2)/30 = v^2.
3000/30 = 100 = v^2, so v = 10m/s, same answer as before.

Hope this helped!