Question: A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to herself and to the bridge (Figure P15.22). The unstretched length

A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to herself and to the bridge (Figure P15.22). The unstretched length of the cord is 11.0 m. She reaches the bottom of her motion 36.0 m below the bridge before bouncing back. Her motion can be separated into an 11.0-m free fall and a 25.0-m section of simple harmonic oscillation.
(a) For what time interval is she in free fall?
(b) Use the principle of conservation of energy to find the spring constant of the bungee cord.
(c) What is the location of the equilibrium point where the spring force balances the gravitational force acting on the jumper? Note that this point is taken as the origin in our mathematical description of simple harmonic oscillation.
(d) What is the angular frequency of the oscillation?
(e) What time interval is required for the cord to stretch by 25.0 m?
(f) What is the total time interval for the entire 36.0-m drop?

A 65.0-kg bungee jumper steps off a bridge

Figure P15.22

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