Climber falls past first protection point toward second

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Homework Statement



Imagine a climber 5.00 m directly abour her uppermost piece of protection, with no slack (exercise lenght) in rope, and the next piece of protection is 6.00 m directly below the uppermost piece of protection. She falls with an acceleration of 9.80 m/s^2 downward. The first piece of protection cuts her velocity in half, but then it fails and she falls farther. What is her speed just when the rope pulls on the second piece of protection?

Homework Equations



deltaX=5m, 6m
a=9.8m/s^2
Vfinal=?

The Attempt at a Solution



Vf^2=Vini^2=2(-9.8)(5)
Vf^2=9.8
9.8/2=4.9

vf^2=Vi^2+2(-9.8)(6)

vf^2=4.9^2+2(-9.8)(6)
Vf=9.64
 

Discussion

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Vf^2=Vini^2=2(-9.8)(5)
Vf^2=9.8
9.8/2=4.9

It's kind of confusing here.
why Vf^2=Vini^2??
and how you got 9.8 there in the second line?
 
well the 9.8 is the acceleration and Velocity initial is from the Velocity final from the 1st equation. I don't know where to go from there.