which formula is this: Vy=sqt(2ay(y-y0)

its used to solve part a of this problem I just don't know where it derives from.
A 75.0-kg man steps off a
platform 3.10 m above the ground. He keeps his legs straight as he
falls, but at the moment his feet touch the ground his knees begin to
bend, and, treated as a particle, he moves an additional 0.60 m
before coming to rest. (a) What is his speed at the instant his feet
touch the ground?

Question

which formula is this: Vy=sqt(2ay(y-y0)

its used to solve part a of this problem I just don't know where it derives from.
A 75.0-kg man steps off a
platform 3.10 m above the ground. He keeps his legs straight as he
falls, but at the moment his feet touch the ground his knees begin to
bend, and, treated as a particle, he moves an additional 0.60 m
before coming to rest. (a) What is his speed at the instant his feet
touch the ground?

rock_mit182 · Answer

\[V _{y}^{2} = V _{yo}^{2} + 2a (y-yo)\ -->  Uniformly accelerated motion along a straight line

rock_mit182 · Answer

In this case, the acceleration vector (a) is constant and lies along the line of the displacement vector (y).

 $$ V _{yo}^{2}=0$$ if there is no initial velocity, the equation will be:

rock_mit182 · Answer

$$V _{y}^{2} =  2a (y-yo)$$

rock_mit182 · Answer

now we take square root in both sides of the equation