Department of Mathematics

IDEA: Internet Differential Equations Activities

Bungee Jumping

This exercise requires the solution of linear second order equations. Maximum and minimum values of parameters must be found to allow a person to survive a jump. The meaning of zeros of the solution and its derivative, as well as the effect of a nonlinear term, are discussed qualitatively.

Suppose that you are a bungee jumper, standing on a bridge somewhere in Colorado. Three hundred feet below you, a lovely little stream meanders through a scenic canyon. The lovely stream is six inches deep, and has lots of sharp pointy rocks in it. You have brought a variety of cords with which to secure your feet, ranging in stiffness from a steel cable to a soft rubber band. Every one of the cords is exactly 160 feet long, when hung from the bridge. If you choose a cord that is too stiff, then your body will no longer form a connected set after you hit the end of the cord. On the other hand, if you choose one that is too soft, your body may still be moving for a short time after you pass the point 300 feet below where you are standing. Which cord should you choose, if any?

This physical system was studied by famous bungee jumper Robert Hooke about 300 years ago. Hooke's Law states that the force exerted on your body by the cord is proportional to the distance of your body past the equilibrium position of the spring. For the present, we will consider all of the mass of your body to be in your feet (the very fact that you are jumping indicates that your head is not very massive). When your body is above the equilibrium position of the cord, then the cord bends, so you may assume that it exerts no force on your body. On the other hand, the force of gravity acts relentlessly on your body. Finally, there is a certain amount of frictional force applied by air resistance. Assume that it is proportional to velocity, and opposes whatever motion is taking place.

The action of these forces may be summarized as follows. Let x = x(t) denote the distance of your body from the end of the bungee cord, i.e. from an imaginary point 160 feet below the bridge. Let the bridge be at x=-160 feet, and the creek be at x=140 feet. Thus x increases as you fall. Define


where k is a constant describing the stiffness of the cord. The larger k becomes, the less stretch there is in the cord. The frictional force has the form tex2html_wrap_inline22. Thus the forces may be summed as


Problem 1: Convert this to a first order system of equations.

Now, you can see that when you come to rest after the jump, your feet will be more than 160 feet below the bridge. After all, gravity is pulling you downward, while the spring exerts no upward force until your feet are more than 160 feet below the bridge.

Problem 2: Find the distance below the bridge that your feet find themselves when you come to rest after the jump for any given cord with stiffness k. Use the knowledge that you are six feet tall and weigh 160 lbs., and the fact that you do not want to rest below the level of the creek, to specify a lower bound on the value for k that you require for your bungee cord. Use tex2html_wrap_inline26. Click here to receive the DynaSys Bungee data file for this problem.

Of course, the simple knowledge that your final rest position lies above the level of the stream may not comfort you that greatly. What you really want to know is how far below the bridge you will fall on your first bounce. After that, the frictional force should take over, and each subsequent bounce should not take you so close to the stream bed. The plot below will allow you to plot your position versus your velocity if air resistance is neglected.

Problem 3: Use the applet and the k value you found in problem 2 to estimate how far you would fall if there were no air.

Problem 4: Solve the differential equation in the case that a=0. You need only to solve from time t=0 until the first time that tex2html_wrap_inline35. Denote this time tex2html_wrap_inline37. When the velocity is zero, that indicates that your body is changing direction, and will go up for a while. Compute the minimum value of the spring coefficient k that will permit you to survive your jump.

Problem 5: When there is no damping of the motion (air resistance), then your trip back upwards after bottoming out will be exactly the opposite of the trip down. Use this fact and the set of axes below to make three plots of the solution curve you have found: one in the (t,x) plane, one in the tex2html_wrap_inline39 plane, and one in the phase plane tex2html_wrap_inline41. Interpret the significance of the intercepts of the curve in the phase plane.

If there were no damping, you would keep bouncing on the end of your cord forever. Fortunately for you, there is air resistance to slow you down. We have already included a linear term for air resistance in the equation above. Such a linear term is unrealistic as a model for air resistance, but we will use it anyway, for the moment. The coefficient is a~1, neglecting such factors as body orientation, clothing, and so on. Since the equation is still linear, you may still solve it.

Problem 6: Solve the differential equation in the case that a=.2. Recompute the minimum value of k that will allow the cord to stop your fall before unpleasant things happen. Test your answer by plotting the solution below.

Let us use a more realistic term for the air resistance. The air resistance is actually a nonlinear function of speed. For this reason, we rewrite the equation as


You will probably not be able to calculate a closed-form solution for this last equation. Instead, use the plot below to compute the solution to the linear equation and that of the nonlinear equation on the same plot. The phase plane appears on the left, a plot of the distance versus time appears at top right, and velocity versus time appears at bottom right. Use a value for tex2html_wrap_inline45 of .005. Click here to receive the DynaSys Bungee data file for this problem.

Problem 7:
Compare qualitatively the solutions for linear air resistance and for nonlinear resistance.

As was mentioned earlier, it is possible for a cord to be too stiff. For example, if you were to jump from the bridge attached to a steel cable, then being stopped by the cable would probably not be significantly better than being stopped by the ground. Suppose we know that a human body tied to a rope can only withstand a pull, from the rope, of about twenty times its weight, beyond which force being caught by the rope is about as bad as hitting the ground.

Problem 8: Use this hypothesis to compute the maximum allowable value for k for a good bungee cord.

Problem 9:Use a computer to draw phase portraits for four of your cords: a collection of rubber bands tied together with k=.2, a bungee cord with k=6, a climbing rope with k=10, and a steel cable with k=300. Which of these is safe? Interpret each plot.

Problem 10: Explain why there are no bungee cords that permit your body to come smoothly to a halt, i.e. you always bounce on the end of the cord.

Other Bungee Links

With the advent of HTML5, Javascript is now ready for prime time for mathematical applications. There are new Javascript demos illustrating how we might use interactive web objects to help students learn Calculus.

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