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h3. Part A

!bungeejump.jpg!
(Photo courtesy Wikimedia Commons, uploaded by user Che010.)

Bungee cords designed to U.S. Military specifications (DoD standard MIL-C-5651D, available at [http://dodssp.daps.dla.mil]) are characterized by a force constant times unstretched length in the range _kL_ ~ 800-1500 N.  Jumpers using these cords intertwine three to five cords to make a thick rope that is strong enough to withstand the forces of the jump. Suppose that you are designing a bungee jump off of a bridge that is 3050.0 m above the surface of a river running 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.)You have read that you should expect the cord to stretch (at peak extension) to about 210% of its natural length.  You have also read that you should use 3 cords together for jumpers with weights in the range 100-150 lbs, 4 cords for 150-200 lbs, and 5 cords for 200-250 lbs.  Suppose you decide to use cords of length 20 m, which would seem to offer a safety zone of about 8 m (or really about 6 m if the cord is attached at the ankles).  

h3.  Part A 
Find the expected maximum length of the cords for a 200 lb person jumping with 4 cords, so that _kL_ = (4)(800 N) = 3200 N.  Since you are findingevaluating athe maximumsafety cord lengthfactor, 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:  

PICTURE

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 "compressed", since they fold like an ordinary rope.  Thus, the initial spring energy is zero in this case.{info}

ThisIt equationseems cannotthat bewe solvedhave withouta furtherproblem constraintshere, sincebecause we do not know _k_h_~f~ or _x_~f~.  The easiest way extrato constraintdeal thatwith wethis haveproblem is to givenutilize byour thefreedom factto that_choose_ the jumperzero haspoint fallenof athe totalheight ofaxis. 28.0 mIf (descendingwe fromrestructure 30.0our mcoordinate abovesystem theto waterplace down_h_ to= 2.0 m above at the point where the water).cord is Thisstretched distanceto mustits benatural coveredlength by the stretched cord.  This gives us the contraint:_L_ (as shown below) then we can rewrite our equation:

NEW PIC

{latex}\begin{large} \[ h_{\rm i}mgL = - hmgx_{\rm f} = L ++ \frac{1}{2} k x_{f}^{2} \] \end{large}{latex}

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

{latex}\begin{large} \[ -2mg(\Delta h) = k((\Delta h)^{2} + 2 L \Delta h + L^{2}) \] \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 haveinfo}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.{info}

We now have a quadratic, which is solved to obtain:

{latex}\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}

whereIt weis havenot replacedsensible thethat quantitywe _kL_should byfind Ca (=negative 800value N) for clarity.  With this substitution, it can be seen that we have a quadratic equation in _L_ which can be solved to findfor _x_~f~, so we must select the plus sign, giving:

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

It is clear{tip}Does this bear out the original estimate 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?cord should stretch to 210% of its initial length?{tip}

{tip}Can you convince yourself by looking at the form of the 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?{tip}

h3. Part B

Based on your analysis from Part A, what would be the peak force acting on a 200 lb person making the jump attached to 4 cords?