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 below.  To get an idea for the maximum cord length, calculate the unstretched length of cord with _kL_ = 4000 N (a system of five 800 N cords) that will result in a 118 kg jumper who leaves the bridge from rest ending their jump 2.0 m above the water's surface.  (The 2.0 m builds in a safety margin for the height of the jumper, so you can neglect the height in your calculation.)  Since you are finding a maximum cord length, ignore any losses due to air resistance or dissipation in the cord.  Ignore the mass of the rope.

System:  The jumper (treated as a [point particle]) plus the earth and the bungee cord.  The system constituents interact via gravity, which contributes [gravitational potential energy], and via the restoring force of the cord, which contributes [elastic potential energy].  External influences are assumed negligible.

Model:  [Constant Mechanical Energy].

Approach:  

Shown above is an initial-state final-state diagram for this situation, along with corresponding energy bar graphs.  As indicated in the picture, we have chosen the zero point of the height to be the river's surface.  With these pictures in mind, we can set up the Law of Change for our model:

{latex}\begin{large} \[ E_{\rm i} = mgh_{\rm i} = E_{\rm f} = mgh_{\rm f} + \frac{1}{2}k x_{f}^{2} \] \end{large}{latex}

{info}Bungee cords provide a restoring force when stretched, but offer no resistance when 

Bungee cords provide an elastic restoring force when stretched, but offer no resistance when "compressed",

since

they

fold

like

an

ordinary

rope.

Thus,

the

initial

spring

energy

is

zero

in

this case.

 case.{info}

This equation cannot be solved without further constraints, since we do not know _k_.  The extra constraint that we have is given by the fact that the jumper has fallen a total of 28.0 m (descending from 30.0 m above the water down to 2.0 m above the water).  This distance must be covered by the stretched cord.  This gives us the contraint:

{latex}\begin{large} \[ h_{\rm i}mgL = - hmgx_{\rm f} = L + x_{f} \] \end{largefrac{1}{latex2}

Solving this constraint for _x_~f~ and substituting into the energy equation gives:

{latex}\begin{large} \[ -2mg(\Delta h) = k((\Delta h) k x_{f}^{2} + 2 L \Delta h + L^{2}) \] \end{large}{latex}

Where we are using Δ_h_ = _h_~f~ - _h_~i~ to simplify the expression.
We still have two unknowns, but we can resolve this by using the fact that _kL_ is a constant for the rope.  Thus, if we multiply both sides by _L_, we have:

{latex}

You can also solve using the initial coordinate system. You would simply have to substitute h_f = 50 m - L - x. After cancelling mg(50 m) from each side, you recover the same expression.

\begin{large} \[ -2mg(\Delta h)L = C((\Delta hx_{f} = \frac{mg \pm \sqrt{(mg)^{2} + 2L2 \Deltak h + L^{2})mg L}}{k} \] \end{large}{latex}

where we have replaced the quantity _kL_ by C (= 800 N) for clarity.  With this substitution, it can be seen that we have a quadratic equation in _L_ which can be solved to find:

{latex}

\begin{large} \[ Lx_{f} = \frac{-2(C+mg)\Delta h \pm \sqrt{4(C+mg)^{2} (\Delta h)^{2}-4C^{2}(\Delta h)^{2}}}{2C} = 13.3 {mg}{k}\left(1+\sqrt{1+\frac{2kL}{mg}}\right) = 21.5\:{\rm m}\;{\rm or}\;58.9 \:{\rm m}\] \end{large}{latex}

It is clear that the appropriate choice is 13.3 m.  

h3. Part B 

Suppose a 118 kg jumper foiled your calculations by leaping upward from the bridge with an initial speed of 2.25 m/s.  Will the jumper hit the water, assuming the 13.3 m cord with _kL_ = 4000 N that we found in Part A?

Does this bear out the original estimate that the cord should stretch to 210% of its initial length?

Can you convince yourself by looking at the form of the symbolic answer that a 150 lb person attached to 3 cords and a 250 lb person attached to 5 cords would have the same x_f as the 200 lb person on 4 cords?