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        riddles >> medium >> Jump! (Ok, but how??)
        
(Message started by: JocK on Jan 24^th, 2005, 3:04pm)

Title: Jump! (Ok, but how??)
Post by JocK on Jan 24^th, 2005, 3:04pm

You are standing with a friend on a cliff. At a depth of 24 m below you is water. Laying next to you is a metal chain. The chain is 24 m long and weights 40 kg. Your friend wants to try his new toy, a laser-doppler velocity meter, on distant moving objects, and - to convince you to jump from the cliff in the sea - offers you a bet: you can earn $ 1,000 if you can hit the water at a speed in excess of 24 m/s. Of course he will use his laser-doppler toy to record your vertical velocity as function of time.

You ask him whether he is prepared to help you a bit. He replies that he will honestly try to do whatever you ask him to do as long as he can keep dry clothes and as long as his precious laser-doppler tool doesn't get damaged.

What do you do to earn $1,000 ?*

* The following applies:

- your own weight is 80 kg.
- you and your friend are both healthy and fit (but neither of you is a super-athlete!)
- the gravitational acceleration is 9.8 m/s[sup2]
- you may assume the air drag to be negligible
- you may assume the chain to be ideal (inelastic but fully flexible)
- the waterdepth exceeds 24 m
- the laser-doppler velocity measurement device is reliable and accurate
- your friend is an honest person
- apart from yourself, your friend, the laser-doppler device and the chain, there is nothing of any assistance to you

Title: Re: Jump! (Ok, but how??)
Post by honkyboy on Jan 24^th, 2005, 9:27pm

You could try [hide]
standing on your friends shoulders and then jumping[/hide]
and/or
[hide]hold on to one end of the chain very tightly (perhaps wrapping it around your waist) and have your friend throw the other end off the cliff[/hide]

Title: Re: Jump! (Ok, but how??)
Post by Noke Lieu on Jan 24^th, 2005, 10:31pm

With out doing any calculations.. will get on that later (busy), am tempted to coil myself in the frictionless chain. The speed of my descent is seemingly increased by the rotation imparted by the unravelling of the chain as I plumet to the 24.1m deep sea.

Title: Re: Jump! (Ok, but how??)
Post by Grimbal on Jan 25^th, 2005, 3:50am

Easy. Jump up 5.5 m.

Seriously, ::[hide]I think tying yourself at one end of the chain and letting the other side slide down would transmit enough downwards momentum to you, before you even start even falling, to make up for the missing height.
But:
- the computations are not simple, so I am not sure,
- you are likely to be hurt when the chain pulls you down,
- you are likely to be hurt again when you hit the water,
- you are going to drown afterwards,
- what kind of friend is this anyway?[/hide]::

Title: Re: Jump! (Ok, but how??)
Post by Eigenray on Jan 25^th, 2005, 2:52pm

m=40kg, M=80kg, d=24m.
Let one end of the chain fall until it's completely straightened (one end on the water, one end on the cliff), and let v be its velocity at that point. By conservation of energy,
mgd/2 = delta potential = delta kinetic = mv²/2,
so v² = gd. If you now grab onto the top end and start falling with it, now at velocity V, then by conservation of momentum,
mv = intial p = final p = (m+M)V,
so V = mv/(m+M), and your energy at this point is
MV²/2 = 1/2 Mm²gd/(m²+M²)
You now fall a distance d, adding Mgd more energy, to get a total of
Mgd[1 + 1/(2(1+r²))]
where r = M/m. To get the final velocity v_f, equate this to 1/2 Mv_f², to get
v_f = sqrt(gd(2+1/(1+r²))).
This only seem to get you to 22.2m/s, for r=2 (as opposed to 21.7m/s for just falling).
Notice that I used conservation of momentum to find V, not conservation of energy. I don't believe kinetic energy would be conserved when the falling chain yanks you over the edge -- for example, if you were holding the chain, and your arm gets pulled away from you as your end of the chain starts moving, the system is doing work on your muscles (which are trying to pull your arm towards you) and therefore loses kinetic energy. But even if kinetic energy were conserved at this point, you'd still only get up to sqrt(gd(2+1/(1+r)))=23.4m/s.

But if you could arrange it so that the chain comes to a stop somehow, transferring its entire mdg kinetic energy to you, you could get up to sqrt(gd(2+1/r))=24.2. Maybe that's where your friend comes in.

As far as coiling yourself in the chain goes, I haven't worked it out but it seems like that would just slow you down, because some of the work done by gravity is going toward rotational energy instead of translational.

Title: Re: Jump! (Ok, but how??)
Post by SWF on Jan 26^th, 2005, 5:10pm

Although I have not carefully checked everything on the numbers, I think the concept would work. Ball up or knot one end of the chain so roughly 1/3 is in the all and 2/3 are still normal chain (it seems to work for a fairly wide range of how much is balled up). Your friend holds the ball end and you hold the other endand pull the chain taut with both of you standing on the edge of the cliff. The friend lets go so the ball swings down like a pendulum with you holding it up. After the pendulum has swung down past 53 degrees, jump off the cliff. As you fall the ball will swing under you accelerating your descent such that you reach the water with sufficient speed- quick estimate shows greater than 26.5 m/s.

It is important not to jump too soon or the ball will strike the water instead of swinging under you. Probably max speed obtained if you strike the water with the chain horizontal, and for an added bonus the chain doesn't fall on you.

Title: Re: Jump! (Ok, but how??)
Post by Sjoerd Job Postmus on Mar 4^th, 2005, 4:27am

Since friction and other factors will not slow you down / prevent you from speeding, we can ignore those. In this situation, not in reality.

We need to hit the water at 24 m/s
v = a * t
24 = 9,8*t
t = 24/9,8 ~= 2.45s

Now, we need to calculate the height we will have to jump off:
x = v_av * t
x = 12 * 2.45 ~= 29,39m

We need to jump from about 29,39 meters. I would use the chain to let me slow down just a tad at the end, because hitting the water would kill me. But, jumping a small 6 meters isn't easy.

We could use the swing effect. If the friend is prepared to hold the chain at a good 7 meters, and let go when we're at the top with the chain straight, we can then fall down from 31 meters instead. (neglecting your friends height)

When you fall down, you get slowed just before the water by the chain tied to your friends waste, and you barely survive. But, we wouldn't reach the water at 24m/s anymore, so what if he just lets go of the chain... we're dead, but he can give the 1000 bucks to someone we care enough about.
---
INVALID: another point (since airdrag is negligible), would to run as fast as you could off the cliff. Airdrag is negligible, so your v_x would remain the same as when you started, your v_y would only reach
x = a * t^2
t^2 = 2.45
v_y = a * t = 15.34 m/s

Now, using pythagoras, we can calculate v_x
24^2 = 15.34^2 + v_x^2
v_x^2 = 24^2 - 15.34^2 = 340.8
v_x = 18,46 m/s ~= 66,46, so we can ignore this.
---

Title: Re: Jump! (Ok, but how??)
Post by THUDandBLUNDER on Mar 4^th, 2005, 9:04am

Quote:

We need to hit the water at 24 m/s

We need to jump from about 29,39 meters. I would use the chain to let me slow down just a tad at the end, because hitting the water would kill me.

Golden Gate Bridge, the favourite suicide spot in US, has a drop of 240 feet and an impact velocity of 75 mph (132 ft/sec or about 40 m/sec). Yet some people manage to survive in spite of themselves - one loser has even survived twice! It is said that the best chance of surviving is when the person enters the water feet first. In fact, a few years ago a stuntman jumped with heavy weights attached to his feet. Unfortunately, a gust of wind flipped him over just before he hit the water and he died.

In your opinion, what is the least height such that, of 1000 Al Qaeda terrorists dropped into the water, no less than 500 will enjoy their primitive idea of Heaven, 70 virgins? ;)

Title: Re: Jump! (Ok, but how??)
Post by Sjoerd Job Postmus on Mar 4^th, 2005, 11:38am

on 03/04/05 at 09:04:16, THUDandBLUNDER wrote:

I stand corrected, no need to survive using these materials... However, I still think the chain would make a great sling for getting really high... ( if lucky, it could even be used to sort of shoot myself off into the sky ( with the risk of falling down on the land instead, but I'll take that risk.

on 03/04/05 at 09:04:16, THUDandBLUNDER wrote:

In your opinion, what is the least height such that, of 1000 Al Qaeda terrorists dropped into the water, no less than 500 will enjoy their primitive idea of Heaven, 70 virgins? ;)

Tough(and offtopic), but I'll see...
3*10^infinity lightyears: They will drop from heaven, so they would all be there already. All 1000 are there.
Another option would be that the group of 1000 included 70 female virgins. Job done. (Wait, you didn't say 70 EACH)

Another point is, that if you have those 1000 dropping from REAL high, they would hit the water with such an energy, that a tidal wave would be created, flooding all major cities. Killing millions! Ok, maybe you would need millions to actually create that wave as well...

Don't know the answer to this :(

Title: Re: Jump! (Ok, but how??)
Post by THUDandBLUNDER on Mar 4^th, 2005, 12:27pm

Quote:

Don't know the answer to this

Maybe there's a knack (http://home7.swipnet.se/~w-72327/quebrada.htm)to it.

.

Title: Re: Jump! (Ok, but how??)
Post by SMQ on Mar 4^th, 2005, 1:17pm

I, also, haven't gone rigorously through the math, but I think SWF was on the right track up there. Here's my solution:

1) You and your friend stand at the edge of the cliff holding the chain as taughtly as possible between you by the ends at waist level

2) Your friend throws his end of the chain downwards at ~6m/s -- maybe just possible with a full-force throw

3) As the chain nears the bottom of its arc its center is traveling at ~10m/s, and the bottom end at ~20m/s. As this point nears you begin to run toward your friend (along the edge of the cliff) at ~4m/s, converting the pendulum motion to rotation about a point 4m from your end of the chain

4) As the chain passes through the bottom of its arc you jump from the edge of the cliff (still traveling at 4m/s nearly parallel to the edge) The system of you and the chain (rotating about its center of gravity 4m from you) begins to fall under the influence of gravity

5) Since you were holding the chain at waist level the CoG of the system falls 21m in just over 2 seconds reaching a speed of just over 20m/s

6) in those 2 seconds the system's rotation carries you through a bit over 90 degrees of arc, converting your 4m/s horizontal motion to 4m/s vertical motion for a total speed of 24m/s when hitting the water.

The exact speed and timing of this manuver would be a bit tricky--most especially you would probably have to jump somewhat *upward* from the edge of the cliff when the chain was passing through ~70 degrees from horizontal on its way down (since in 2 seconds a 4m/s velocity on a 4m radius carries you through 110 degrees of arc)--but with proper planning and a friend strong enough to give the chain a sufficient initial velocity, I think it should work!

--SMQ

Title: Re: Jump! (Ok, but how??)
Post by Octavian on Apr 29^th, 2006, 9:17am

on 03/04/05 at 09:04:16, THUDandBLUNDER wrote:

In your opinion, what is the least height such that, of 1000 Al Qaeda terrorists dropped into the water, no less than 500 will enjoy their primitive idea of Heaven, 70 virgins? ;)

You definetely haven't got enough bus-bombing, brit??
Coming more soon!

Title: Re: Jump! (Ok, but how??)
Post by JocK on May 27^th, 2006, 12:45pm

Guys, I'm gonne give you a (pretty strong) hint:

[hide]What happens if you secure one end of the chain to your waist, ask your friend to hold the other end, lower the middle part of the chain such that it hangs down the cliff, and then jump...? [/hide]