The Royal Gorge bridge over the Arkansas River is 310 m above the river. A 65-kg

Question

The Royal Gorge bridge over the Arkansas River is 310 m above the river. A 65-kg bungee jumper has an elastic cord with an unstressed length of 57 m attached to her feet. Assume that, like an ideal spring, the cord is massless and provides a linear restoring force when stretched. The jumper leaps, and at at her lowest point she barely touches the water. After numerous ascents and descents, she comes to rest at a height h above the water. Model the jumper as a point particle and assume that any effects of air resistance are negligible.

Explanation / Answer

With outstretched arm, the lady might be supposed to be about 2m tall. However you don’t mention this as a consideration so, not seeking unwanted refinements, I shall ignore her stature and assume her to be a mere “point” at the end of the cord! So the cord extends all the way to 310 m.

Potential energy of lady at takeoff = energy in cord as she touches the water = 65 x 9.81 x 310 = 197671 J.
Consider the fall to start at x=0 and finish at x= 310
Energy entering cord as the lady moves from x to x+dx = K(x-57).dx where K is the elastic constant of the cord in N/m extension.
So total energy stored in cord during descent = K.(x-57).dx (between limits x=57 to x=310)
= [x²/2 – 57x] (from x=57 to x=310) = K 29760. = 19767. giving K = 5.622 N/m

At h above the water the cord is extended an amount E. We can write KE = 65 x 9.81 = 637.65 N. Inserting the above value we found for K immediately yields:
E = 637.65/5.622 = 113.42 m which makes the height h = 310 – 57 – 113.42 = 140 m

At the point of maximum velocity, force in cord = weight of lady (acceleration = 0)
KE’ = 65 x 9.81 giving E’ = extension of cord at this point = 65 x 9.81/5.622 = 113.42m
Energy in cord at this point = K(E’²/2) = 5.622 x 113.42 /2 = 31882 J
So lady’s kinetic energy = 197671 – 31882= 165789 J
So lady’s speed v = (2 x 165789 / 65) = 71.4 m/s

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