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KURD86
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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