Problem — Bungee jump

Problem

A person ($80\;\mathrm{kg}$) jumps from a platform $100\;\mathrm{m}$ high with a bungee cord ($K = 16\;\mathrm{N/m}$, $\ell_0 = 0$). How deep is the lowest point of the jump? Set up the balance between gravitational and elastic potential energy and use the geometric constraint $h_B + \ell_B = 100\;\mathrm{m}$.

Solution

Setup. We define two states:

• State A (start): the person is at rest on the platform, $v = 0$, the cord is slack ($\ell_0 = 0$, so $E_\text{pot,m} = 0$).
• State B (lowest point): the person is again at rest ($v = 0$, maximum stretch of the cord), the cord is stretched by $\ell_B$ and the person is at height $h_B$ from the ground.

Since $\ell_0 = 0$, the cord begins to stretch as soon as the person leaves the platform: its length at the lowest point coincides with the distance fallen. The geometric constraint links the two heights:

$h_B + \ell_B = 100\;\mathrm{m}$

Energy balance (no friction, $v = 0$ at both A and B). The gravitational fall converts all of its energy into elastic energy of the cord. Taking the ground as reference:

$mg\cdot(100) = mg\cdot h_B + \tfrac{1}{2}K\,\ell_B^2$

Since $100 - h_B = \ell_B$, the net gravitational term is $mg\,\ell_B$:

$mg\,\ell_B = \tfrac{1}{2}K\,\ell_B^2$

Quadratic equation in $\ell_B$:

$\tfrac{1}{2}K\,\ell_B^2 - mg\,\ell_B = 0 \quad\Rightarrow\quad \ell_B\left(\tfrac{1}{2}K\,\ell_B - mg\right) = 0$

The solution $\ell_B = 0$ is the starting one; the physical one is:

$\ell_B = \frac{2mg}{K} = \frac{2\cdot 80\cdot 9{,}81}{16} = \frac{1569{,}6}{16} \approx 98{,}1\;\mathrm{m}$

Result. The cord stretches by about $98{,}1\;\mathrm{m}$: the lowest point is therefore at

$\ev{\ell_B \approx 98{,}1\;\mathrm{m}\ \text{below the platform}, \qquad h_B = 100 - 98{,}1 \approx 1{,}9\;\mathrm{m}\ \text{above the ground}}$

The person stops just $1{,}9\;\mathrm{m}$ above the ground: a well-tuned jump. Note that at the lowest point the velocity is zero but the acceleration is not — the elastic force ($K\ell_B \approx 1570\;\mathrm{N}$) exceeds the weight ($mg \approx 785\;\mathrm{N}$), so the person is pulled back upwards.

Collegamenti

Argomenti: Work and energy Concetti: Gravitational potential energy · Elastic potential energy · Conservation of mechanical energy · Elastic force and Hooke’s law Competenze: Conservation of energy · Symbolic setup Oggetti: Spring