Falling in an arc

Factual Questions
1

Given: Objects near the surface of the Earth fall at a rate of 30 feet per second, per second.

Given: A canyon 30 feet wide and more than 30 feet deep.

Given: A rope 30 feet long, one end of which is fixed to the rim of the eastern wall of the canyon. Assume the rope is entirely inelastic.

Given: You, attached by harness to the other end of the rope, holding on to the rim of the western wall of the canyon.

Disregard: The effects of wind or air resistance.
You let go of the western rim. How long will it take you to hit the side of the eastern wall 30 feet down and 30 feet away? How fast will you be moving when you hit it? And will it make you die?

2

My guess would be just under 30 mph. (28.93 mph)

I’m probably wrong.

3

Could you show your math? Because even if you weren’t tethered, you’d fall 30 feet in 1 second, which is only 20 miles per hour; much less than 30 mph or even 28.93 mph.

4

Fiver,

  1. Lance Turbo is correct, given your numbers (the acceleration of gravity near the Earth’s surface is actually closer to 32 ft/s[sup]2[/sup]).

  2. Don’t confuse acceleration and velocity. In free fall, you only fall 15 feet in the first second.

  3. To solve the problem, use conservation of energy.

  4. I’m not giving any more details, in case this is homework.

Rick

5

RickG, will you give details? I assure you Fiver is a grown-up and the only classes currently enrolled are tap-dance.

Thanks, ‘cause I’m curious too.

6

Fiver, Why did you let go of the western rim? Don’t you know the result is gonna’ hurt?

(Will get back later with the math stuff)

7

E=1/2 mv[sup]2[/sup] + mgh.

Let your impact point, 30 ft down from the top of the canyon be h=0. g=30 ft/s[sup]2[/sup]

At the instant you let go, your v=0, and h=30, so E=mgh=m*900 ft[sup]2[/sup]/s[sup]2[/sup].

When you hit, your h=0, so your energy is E=1/2 m v[sup]2[/sup]. Energy is conserved, so 1/2 m v[sup]2[/sup]= m * 900, so v=42.43 ft/s.

The time part is tougher. Involves calculus and stuff. I’ll post it tomorrow, if nobody has by then.

8

No takers, eh? What, yer scared of Legendre’s elliptic integrals?

Basically, Fiver wants to be the bob on a pendulum released from an angle of ninety degrees. Physics 101 veterans know that it’s easy to solve the pendulum problem if we can use the small-angle approximation, sin(theta)=theta. However, ninety degrees isn’t a small angle, so we are forced into the mathematical the ugliness alluded to above, which gives us the period of the pendulum: 7.4164sqrt(l/g), or 7.4164sqrt(30 ft/ 30 ft/s[sup]2[/sup]), or 7.4164 seconds. Obviously, due to that big mean wall getting in the way, Fiver will only execute one fourth of a period, so his fall will last 1.8541 seconds.

Not much of a thrill ride, if you ask me.

9

Podkayne did a great job following up with the details. Thanks, Podkayne!

Rick

10

But, you ask, is 1.845 seconds enough time to see my life flash before my eyes? In other words, the last question,

Will it make you die?

Probably yes. Try riding your bicycle into a wall at 30 mph (make sure you’re not clipped in). While someone might know a little more about this than me, we can pretty much only guess how quickly you’ll be stopping. This, and your mass, determines the amount of force you’ll receive from the wall when you die.

The impulse (change in momentum) for a 150-lb person going from 42.43 ft/s to 0 ft/s is (42.43 * 150 ) 6360 ft-lb/s. This should actually be in ft-slug/s, to get 200 lbf-s. I’m switching to SI anyway.

Using the heavier 70-kg SI human, the impulse is 900 N-s. Your velocity on impact is 12.93 m/s . Somewhere around a 12 kN force is what I’ve heard will kill you. To get average force, we need an estimate of time.

If you go splat like Wile E. Coyote (i.e. the wall stops you completely), I’d guess your body would stop in something like 10 mS. The force of this is 90 kN, easily enough to kill you and make a not-so-pretty cliff ornament.

More likely than not, though, you’ll bounce. If you bounce, the impulse is actually larger but the amount of time will certainly be longer. So if you bounced back high enough to swing 1 m up, the impulse is now 1200 N-s. But maybe you took more than twice as long, say 25 ms. This gives a force of about 50 kN. Still enough to kill you, but it’s a bit more pleasant to look at for the rest of us.

[diversion on climbing ropes : static vs. dynamic]
Now for something else : Suppose you just fell of the east side of the canyon, not the west side. In other words, straight down. Do you still die? Of course, it just happens sooner. Your rope is inelastic, so you get the same impulse in about the same amount of time.

Let’s say your rope is slightly elastic; it only stretches 0.5%. That’s 2 in. on 30 ft. You go from full speed to stop in this distance, and the average force is 130 kN. Your organs are mush.

If, however, you were using a normal rock-climbing rope (a dynamic rope), you might expect it to stretch about 7% (manufacturers will specify). This is 25 in. on 30 ft. You take about 100 ms to stop, and experience a force of only 9.8 kN. This is serious, but probably not deadly. At any rate, you’ll actually be bouncing on this rope, so the force will be a lot less.

The point of this is, don’t go climbing on cheap ropes. And if you’re on a static rope, make sure you’re not going to fall.

[/diversion on ropes]

11

On whether it would kill you:

The math above is good, but a lot of assumptions have been worked in … an easier way to look at it is by remembering that the impact speed is exactly the same as from a straight 30 foot fall. There are some small differences-you’ll have more time to prepare for impact, for instance-but in essence it’s the same type of impact. And a 30-foot drop onto a hard surface will quite likely kill you, or at least break limbs and rupture internal organs.

Moral of the story: do not under any circumstances try to imitate the rope bridge stunt in ‘Indiana Jones and the Temple of Doom’. :slight_smile: