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If you use a steel cable instead of a bungee cord to jump off a bridge, then you would hit the end of the cable at a greater impact than you would with something made of elastic which forces your back up at the end or point of impact. It is all about BOUNCING!

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Q: 6 Why is it a serious folly for a bungee jumper to use a steel cable rather than an elastic cord?
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Why is it a serious folly for a bungee jumper to use a steel cable rather than an elastic cord?

The answer: Small stretch means small stopping time means greater force. Ouch!


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What forces act on a bungee jumper no advanced calculus?

Without a calculus explanation; you can say that the only forces that act on a bungee jumper is F(gravity), and F(rope), and friction. F(gravity) is the force of gravity that pulls the jumper down F(rope) is the force of the rope that is tension that will later keep the jumper from slamming into his death and friction, due to the air. The formula for free fall acceleration is delta y = (vinitial)(t) + (1/2)g(delta t)^2 (which only happens when, hopefully never happens, the bungee rope breaks and you are no longer attached, yikes!). Aww, no "advanced" calculus? Maybe a little basic calculus, or rather algebra, or none at all. The F(rope) mentioned above is not actually a constant force, since it results from the elastic bungee attached to the jumper. The force is actually proportional to the distance it is stretched: F = -kx, negative being that the force is in the opposite direction of the displacement (i.e. you pull a spring left, the force pulls it back to the right). Since the force is not constant throughout the entire "trip" the force you experience is not constant, resulting in higher "g" force at the bottom of the fall, and instantaneous free fall at the top of the recoil. Ok, maybe a little calculus to satisfy me and other people who may be interested in the Calculus. There is still time to hit the back button. Alrighty, Since you are falling, there is a change in potential energy due to gravity: delta U. Since U = -W, and W = integral (F x dx ), W = integral (-kx dx) (from above equation), which integrates to (1/2)kx2, therefore U = -(1/2)kx2. Using this equation, you can find the period of oscillation for this dampen oscillation (since energy is lost to heat and gravity does work on the system). That wasn't so bad was it?


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