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# What is the ball's height at the lowest point of its trajectory?

Consider a bungee cord of unstretched length

L0 = 48 m. When the cord is stretched to

L > L0 it behaves like a spring and obeys

Hooke’s law with the spring constant k =38 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 = 138 m above the

surface of a river) and the other end is tied to

a steel ball of weight mg = 84 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

What would the upward acceleration of the ball at the lowest point be?

### 1 Answer

- NCSLv 71 month agoFavorite Answer
The ball's GPE is converted into EPE in the "spring":

m*g*(H - hbot) = ½kx²

where x = spring extension = H - hbot - L0

so

84kg*9.8m/s²*(138m - hbot) = ½*38N/m*(138m - hbot - 48m)²

This quadratic has solutions at

hbot = 119 m ← which is actually "htop"

and hbot = 17.8 m ≈ 18 m ◄ solution

Hope this helps!

At the lowest point, the net force is

F = kx - mg = 38N/m * (138-18-48)m - 84kg*9.8m/s²

F = 1913 N

a = F / m = 1913N / 84kg = 22.8 m/s² ≈ 23 m/s²