I have the following problem:

The Royal Gorge bridge over the Arkansas River is $310\text{ m}$ above the river. A $57\text{ kg}$ bungee jumper has an elastic cord with an unstressed length of $64\text{ m}$ attached to her feet. Assume that, like an ideal spring, the cord is massless and provides a linear restoring force when stretched. The jumper leaps, and at at her lowest point she barely touches the water. After numerous ascents and descents, she comes to rest at a height h above the water. Model the jumper as a point particle and assume that any effects of air resistance are negligible.

(a) Find $h$.

(b) Find the maximum speed of the jumper.

I was able to solve part (a) (the answer is $148.3\text{ m}$), but I can't figure out how to do part (b). How do I go about solving it?

(As an aside, I calculated that the spring constant for the bungee cord is $k=5.72$.)

One thing I tried was setting the gravitational potential energy at the top of the jump equal to total energy when the jump was h meters above the water:

$U_{grav} = K_{grav} + U_{spring} + U_{grav}$

$mg(310) = \frac12mv^2 + \frac12kh^2 + mgh$

$(57*9.8*310 = \frac12(57)v^2 + \frac12(5.72)(148.3)^2 + 57*9.8*148.3$

This gives me $v=31.02\text{ m/s}^2$, which is incorrect.

  • $\begingroup$ One thing that I just noticed is that 148.3 m is the distance up from the bottom of the drop, not down from the top of the drop. Correcting for this mistake doesn't get me the right answer, though, so my approach is still wrong... $\endgroup$
    – Kevin
    Feb 27, 2013 at 6:46
  • $\begingroup$ ah what is the corect answer $\endgroup$
    – Deiknymi
    Feb 27, 2013 at 7:06
  • $\begingroup$ @Akash I don't know. This problem was given to me in an on-line homework assignment, and it tells me when I'm wrong, but doesn't let me know what answer I should get >_< $\endgroup$
    – Kevin
    Feb 27, 2013 at 9:51

2 Answers 2


When the person jumps, he will be freely falling until the cord reaches its unstretched length $\ell_0$, so during this entire time, the speed of the jumper will be increasing. Additionally, even after the cord starts to stretch, the speed of the jumper will increase until the cord stretches enough that its restoring force equals the force due to gravity. When this happens, the acceleration will be zero, the speed will be a maximum (because it has been increasing the whole time up to this point), and immediately after this point, the speed will start to decrease. Mathematically, you can calculate the distance below the initial height that this point corresponds to by setting the restoring force of the cord equal to the force due to gravity $$ k\Delta\ell = mg $$ and then adding $\Delta \ell$ to the unstretched length $\ell_0$. So $\ell_0 + \Delta \ell$ is the distance below the bridge at which the speed is at a maximum.

Now use conservation of energy to determine the speed of the jumper at this distance below the bridge!

To do this, let's set the zero of potential energy at the bridge, then the potential and kinetic energies of the jumper are both zero just before the jump. However, at a distance $\ell_0+\Delta\ell$ below the bridge, the gravitational potential energy with be $-mg(\ell_0+\Delta\ell$, the spring potential energy will be $\frac{1}{2}k\Delta\ell^2$, and the kinetic energy is unknown and equal to $\frac{1}{2}mv^2$ where we want to solve for $v$. Putting this all together and using conservation of energy gives $$ 0 = -mg(\ell_0+\Delta\ell) +\frac{1}{2}k\Delta\ell^2+ \frac{1}{2}mv^2 $$ and now you can solve for $v$.

  • $\begingroup$ To be honest, I don't understand how to use conservation of energy to solve the problem. Can you give me the first couple of steps in that part of the solution? $\endgroup$
    – Kevin
    Feb 27, 2013 at 9:53
  • 1
    $\begingroup$ I edited the response to include more detail. $\endgroup$ Feb 27, 2013 at 17:04

Well iam not giving the answer but solving the problem according to he method suggested by joshphysics. Here's the solution

                              k*x = mg
                               x  = mg/k
                               x  = 57*10/5.72
                               x  = 99.65m

so now apply the energy conservation

                              mgh = mg(l+l') + (mv^2)/2 

hope you can solve further and find the value of speed.


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