Question

Consider a bungee cord of unstretched length L0 = 45 m. When the cord is stretched to L > L0 it behaves like a spring and obeys Hooke’s law with the spring constant k = 33 N/m. However, unlike a spring, the cord folds instead of becoming compressed when the distance between its ends is less than the unstretched length: For L < L0 the cord has zero tension and zero elastic energy. To test the cord’s reliability, one end is tied to a high bridge (height H = 214 m above the surface of a river) and the other end is tied to a steel ball of weight mg = 150 kg × 9.8 m/s 2 . The ball is dropped off the bridge with zero initial speed. Fortunately, the cord works and the ball stops in the air before it hits the water — and then the cord pulls it back up. Calculate the ball’s height hbot at the lowest point of its trajectory. For simplicity, neglects the cord’s own weight and inertia as well as the air drag on the ball and the cord. Answer in units of m.

Answer 1

Answer:

   y1 = -130 m

            y2 = 130.99 m

Explanation:

For this exercise we can use energy conservation

Starting point Higher

     $Em_{o}$ = U = mg h

     .h = 215 m

Final point. In the lowest part of the trajectory

      Em_{f} = Ke + U

      Em_{f} = ½ k (y-45) 2 + m g y

Energy is conserved so

           Em_{o} = Em_{f}

           m g h = ½ k (y- 45)² + m g y

           150 9.8 214 = ½ 33 (y-45)² +150 9.8 y

           314580 = 16.5 (y² - 2 45 y + 45²) + 1470 y

           314580-16.5 2025 = 16.5 y² - 1485 y + 1470y

            281169.5 = 16.5 y² - 15 y

            y² - 0.9090 y - 17040.57 = 0

We solve the second degree equation

            y = [0.9090 ±√ (0.9090² + 4 17040.57)] / 2

            y = [0.9090 ± 261.08] / 2

            y1 = -130 m

            y2 = 130.99 m

Therefore the body drops to 130.99 m