Rock Climber problem

1. Sep 14, 2007

KURD86

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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

2. Sep 14, 2007

rootX

It's kind of confusing here.
why Vf^2=Vini^2??
and how you got 9.8 there in the second line?

3. Sep 14, 2007

KURD86

well the 9.8 is the acceleration and Velocity initial is from the Velocity final from the 1st equation. I dont know where to go from there.