Elastic Energy Problem...NEED HELP!!?

Question

An 86.0 kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 x 10^3 N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope streched when it breaks his fall and momentarily brings him to rest?

My Approach:
1/2(86)(3.38 m/s) + (86)(9.8)(0) = 1/2(86)(0) + 1/2(1200)x --- i solved from the middle and the bottom

145.3 = 600x

x = .242 m

I found PE and KE at the middle and PE(elastic) and PE at bottom...the answer is wrong for sure though--it doesn't even come close to the book answer at 1.95m

Answer 1

The climber falls through a height h (= 0.75 m)

h = 0.5*g*t^2

Solve for t, the fall time before the rope affects the motion. At the end of that time the climber's speed is

v = g*t

At the point that the rope runs out of slack, the climber's total energy is

m*g*y + 0.5*m*v^2

When the rope runs out of slack it stretches, behaving like a spring, so that when the climber stops falling (briefly) his total energy is now

m*g*(y - x) + 0.5*k*x^2

where x is the distance the rope stretched, and (in both the above expressions) y is the height above the ground. Consevration of energy says

m*g*y + 0.5*m*v^2 = m*g*(y-x) + 0.5*k*x^2

Simplifying this leads to

0.5*m*v^2 = -m*g*x + 0.5*k*x^2

and rearranging gives

x^2 - (2*m*g/k)*x - m*v^2/k = 0

This quadratic equation can be solved for x, the distance through which the rope stretched. Note: be sure to take the physically consistent root when solving the quadratic equation.

Answer 2

Actually you can lump the free-fall distance with the input PE rather than compute a separate KE term. Then you have:
.5*k*x^2 - m*g*(x + .75) = 0 which, grouped by powers of x, is
.5*k*x^2 - m*g*x - m*g*.75 = 0
The positive root is 1.947 m.
Asker's approach erred in first (KE) term treatment of v (should be 3.835 m/s, and should be squared); 2nd (PEgrav) term should be (86)(9.8)(x); and last (PEelastic) term .5*1200*x^2. That way you end up with a quadratic as answerers have pointed out. Get your PE and KE equations straight in your head!