Question

A stuntman wants to bungee jump from a hot air balloon 60.0 m above a the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. Model his body as a particle and the cord as having negligible mass that obeys Hooke's law. In a preliminary test, hanging at rest from a 5.00 m length of the cord, he finds that his body weight stretches it by 1.30 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. Hint: the spring constant changes with a longer piece of cord.

A) What lenght of cord should he use?
B) What maximum acceleration will he experience?

Answer 1

A) 24.7 m, B) 28.9 m/s²

Step-by-step explanation:

Hooke's law states the force of a spring is equal to the spring constant times the change in length:

F = k ΔL

Solving for k:

k = F / ΔL

The spring constant is inversely proportional to the length:

k ∝ 1/L

Therefore:

k₁ L₁ = k₂ L₂

(F₁ / ΔL₁) L₁ = k₂ L₂

(mg / 1.30) (5.00) = k L

k = (5.00/1.30) (mg / L)

Initial energy = final energy

Initial gravitational energy = final gravitational energy + elastic energy

mgH = mgh + 1/2 k (ΔL)²

mg(H - h) = 1/2 k (ΔL)²

mg(60.0 - 10.0) = 1/2 k (ΔL)²

50mg = 1/2 k (ΔL)²

100mg = k (ΔL)²

The stuntman will fall a distance L and then an additional distance ΔL. We know this distance is equal to 60-10 = 50 m. L + ΔL = 50, so ΔL = 50 - L.

100mg = k (50 - L)²

100mg = k (2500 - 100L + L²)

100mg = (5.00/1.30) (mg / L) (2500 - 100L + L²)

26L = 2500 - 100L + L²

0 = L² - 126L + 2500

L = (126 ± √5876) / 2

L = 63 ± √1469

L ≈ 24.7 m, 101 m

Obviously L can't be more than 50.0 m, so L = 24.7 m.

As a mass on a spring, the stuntman will follow simple harmonic motion, so his maximum acceleration will be experienced at his minimum velocity, or at the very bottom.

∑F = ma

k ΔL - mg = ma

(5.00/1.30) (mg / L) ΔL - mg = ma

(5.00/1.30) (g / L) ΔL - g = a

(5.00/1.30) (9.81 / 24.7) (50.0 - 24.7) - 9.81 = a

a = 28.9 m/s²