Dvs Descendant

Introduction
Recently, a consultant to the Science Fair sent me a message asking about a solution proposed in the college entrance exam by FURG 2007. Here is the proposed topic:

"An adventurer plans to jump off a bridge tied to a rubber band (a radical sport known for" bungee jumping) cable. The other end of the cable tie in the bridge. Initially, the movement of the rider is a free fall. Beginning at the point in that the cable is taut, the rider starts to slow down to a certain position where it stops. From that moment forward, the cable starts to pull the jumper on the rise. This position, where the rider will reverse the fall, marking his / her greatest vertical displacement D on the bridge. Of course the height of the bridge should be larger than D. He considers a 80 kg rider weight hypothetical situation at this time on a bungee cord 20 m length. The elastic constant of the cable is 160 N / m. Calculate the value of D.
Comment: the mass of the cable can be disregarded with respect to the mass of the jumper. For an acceleration of gravity, using the value of 10 m / s ²
L ') 20 m. B) 25 m. C) 40 m. D) 36 m. And) 10 m. "

Be landed on the search engines with discriminador "bungee jump" is made, that the return of hundreds (or thousands) of places that you / they say everything about the sport, less like him, he --- This really works and is the objective of the proposed topic exactly. This article aims to fill this gap and, as it is science, all suggestions and criticisms are quite right.

I will always keep in view the solution of the proposed topic and the possible routes for you to solve it. Like, let us improve the data of the object:

rider weight: m = 80 kg
natural length of cable: L = 20 m
Cable elastic constant k = 160 N / m
local acceleration of gravity g = 10m / s ²
(To neglect the effects of air and ground cable)

We should determine to what distance D will remain the rider from the bridge.

First Solution
In this direction, I will first follow the technique of Newton's laws, to check the active forces in the system. In a second solution that we use the conservation of energy.

Referencial and phases of the process
Adopt as reference axis Y in the vertical plane containing the point and jump in the bridge that we take y = 0; the positive direction is down ". The "leap" has two different phases.
The first freefall is checked (grandmother = 0) because the cable is not always required, so that for some time interval, in other words, while y <L, the agent of the strength in man (hence resulting) is its own weight and its acceleration is g = constant and the speed of an object is in freefall.

Equacionamento the first phase

Y <L. ...... F = mg = P ..... l '= g. ..... v ² = 2.gy. ...... (1)

In this phase (1) ..... the g => 0, v> 0 ..... the V and have the same sign (the vectors g and v are two vertical down, same direction Y), we have a uniformly accelerated motion.

When the distance from the man on the bridge, reached y = L, was considered the second phase: the bungee cord begins to be distorted (to stretch) and grape applying a retraction force (elastic Hooke's Law, f =- ks) type of man (opposite F f). Like this, for y> The present cable deformation s = y - L, so that:

Y> L. ..... f = - = KS - k. (Y - L )...... (2)

Man is represented by point "equipment" (he / she plans black) of gravity (weight) force to the blue vectors and elastic (f) force to the cable to the red vectors. To "help" the presentation, the diagrams are.