A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will

Question

A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 m above the ground.  Model his body as a particle and the cord as having negligible mass and obeying Hooke's law.  In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.75 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.     (a) What length of cord should he use?
          m

    
    (b) What maximum acceleration will he experience?
          m/s2




A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 m above the ground.  Model his body as a particle and the cord as having negligible mass and obeying Hooke's law.  In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.75 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.     (a) What length of cord should he use?
          m

    
    (b) What maximum acceleration will he experience?
          m/s2



    (a) What length of cord should he use?
          m

    
    (b) What maximum acceleration will he experience?
          m/s2
A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 m above the ground.  Model his body as a particle and the cord as having negligible mass and obeying Hooke's law.  In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.75 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.     (a) What length of cord should he use?
          m

    
    (b) What maximum acceleration will he experience?
          m/s2

Explanation / Answer

Weight = mg

Stiffness of cord = mg/1.75

energy conservation: Potential energy change = Energy stored in cord

mg(h1 - h2) = 1/2*kx^2

mg*(72-15) = 1/2*(mg/1.75) *(72 - 15 - L)^2

Solving this, L = 42.875 m

b)

max. acceleration = g = 9.81 m/s^2

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