Question:

A bungee jumper of mass m attaches one end of a light elastic string to their foot, and the other end to the railing of a bridge on which they stand. The string has natural length l₀ and modulus of elasticity λ.

(i) Write down the energy equation.
(ii) Use conservation of energy to determine the maximum distance the jumper descends after stepping off the bridge.
Hint: Set the origin of the gravitational potential energy at a distance l₀ below the bridge (i.e., where the string just starts to stretch).

Answer:

Step 1: Define Variables and Set Energy Reference

Let x be the extension of the string beyond its natural length l₀. So the total fall distance becomes l₀ + x.

At the Jumping Point (Initial State):

• Gravitational Potential Energy: 0 (chosen as reference point)
• Kinetic Energy: 0 (jumper starts from rest)
• Elastic Potential Energy: 0 (string not stretched yet)

At the Lowest Point (Final State):

• Gravitational Potential Energy: -mgx
• Elastic Potential Energy: ½ λx²
• Kinetic Energy: 0 (velocity = 0 at maximum descent)

Step 2: Apply Conservation of Energy

Total mechanical energy is conserved:

Step 3: Solve the Equation

Multiply both sides by 2 to eliminate the fraction:

Factor the equation:

The two solutions are:

Therefore, the total fall distance from the bridge is:
l = l₀ + x = l₀ + (2mg / λ)

Final Answer:

• Energy Equation: E = -mgx + (1/2)λx²
• Maximum Fall Distance: l = l₀ + 2mg / λ

Conclusion:

By applying energy conservation principles, we derived the maximum fall distance for a bungee jumper using an elastic cord. The gravitational potential energy lost by the jumper is exactly converted into elastic potential energy stored in the cord, providing a precise and elegant way to calculate the motion and safety of the jump.