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Its All Physics Baby!


Cruising_Adventuring
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hey everypne i want to do a puzzle cache. Here is an example of a question that i might ask. (if you havent guessed physics is my favorite thing in the world!)...well sorta.

 

 

 

A wrench falls from a helicopter that is rising steadily at +6.0 m/s. After 2.0s

a. what is the velocity of the wrench?

b. how far below the helicopter is the wrench?

 

if you guys could try to solve this it would help me out. Just tell me whether you would do this type of problem to get the coordinates of a cache, or if it takes to much time.

 

ALSO NEGLECT ALL FRICTIONS. (keep it easy)

Edited by Cruising_Adventuring
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It's been a bit, so please excuse me if my physics are off.

 

I'm using g=9.82 m/s^2 for simplicity sake.

 

The velocity would be 19.64 m/s in the downward direction

 

And the wrench would be 19.64 m below the point which it was dropped from (also assuming this is the center of mass for the helicopter for simplicity sake).

 

 

I wouldn't mind doing this beforehand, as long as things (such as what you used for gravity) are given. I'm not sure if it would be widely enjoyed, but I would venture that a fair number of people would enjoy it, as long as they understand the physics behind the question you pose.

 

 

edit: Added response to questions...hehehe

Edited by The Psion
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It's been a bit, so please excuse me if my physics are off.

 

I'm using g=9.82 m/s^2 for simplicity sake.

 

The velocity would be 19.64 m/s in the downward direction

 

And the wrench would be 19.64 m below the point which it was dropped from (also assuming this is the center of mass for the helicopter for simplicity sake).

 

 

I wouldn't mind doing this beforehand, as long as things (such as what you used for gravity) are given. I'm not sure if it would be widely enjoyed, but I would venture that a fair number of people would enjoy it, as long as they understand the physics behind the question you pose.

 

 

edit: Added response to questions...hehehe

i usually use 9.8m/s/s for gravity just to make calculations quicker...but 9.82 is more exact. However the resulting velocity of the wrench would be a negative number to indicate direction. Positives for up, negatives for down. Your first answer is close but not quite right. Also when finding how far below the helicopter the wrench is, take into consideration how high the helicopter was when the wrench dropped, and how how high the helicopter was the exact second the wrench hit the ground when is is accelerating at -9.8m/s/s. These two need to be added together because the helicopter is moving +6.0m/s even afer the wrench is dropped.

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If I were to take said wrench that hit the ground after it was dropped from the helicopter grasping it at its midpoint with jaws forward and throw it in a forward overhand motion towards your general direction from a distance of 15 feet:

A - the wrench would strike your forehead at what velocity?

B - the wrench would imbed how many inches?

 

I’m using frustration factor of 10.0. Just trying to be sarcastic here, would I be able to figure this out in time to grab a FTF - NO, would I try to figure is out - YES.

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If I were to take said wrench that hit the ground after it was dropped from the helicopter grasping it at its midpoint with jaws forward and throw it in a forward overhand motion towards your general direction from a distance of 15 feet:

A - the wrench would strike your forehead at what velocity?

B - the wrench would imbed how many inches?

 

I’m using frustration factor of 10.0. Just trying to be sarcastic here, would I be able to figure this out in time to grab a FTF - NO, would I try to figure is out - YES.

I got a good laugh outta that one. lol

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The velocity would be 19.64 m/s in the downward direction

 

Your right i didnt see the downward direction part. ( i was also thinking of making my sample problem an adding vectors problem but choose this one instead)

 

However, Those figures are incorrect. real close but your wrench velocity is incorrect. But looking answer b it looks like you added the figures up correctly.

Edited by Cruising_Adventuring
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If I were to take said wrench that hit the ground after it was dropped from the helicopter grasping it at its midpoint with jaws forward and throw it in a forward overhand motion towards your general direction from a distance of 15 feet:

A - the wrench would strike your forehead at what velocity?

B - the wrench would imbed how many inches?

 

Well for a start....

 

D=15 feet = 4.6 meters

 

A=?

 

V=?

 

Imbed= how hard my skull is?

 

lol

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Sure, I would probably go for a cache like this. As mentioned, it would be helpful if the constants were specified (I would have used g=9.81 m/s^2, just because that is what I have always used).

 

Methinks, there is an error in your last formula. It should read:

 

Vf^2 = Vi^2 + 2ad

 

All the best for a great puzzle cache. It has given me an idea for a cache of my own.

 

Cheers

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If I were to take said wrench that hit the ground after it was dropped from the helicopter grasping it at its midpoint with jaws forward and throw it in a forward overhand motion towards your general direction from a distance of 15 feet:

A - the wrench would strike your forehead at what velocity?

B - the wrench would imbed how many inches?

 

:lol::lol::lol::lol:

 

Roaring with laughter!!!!!

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I've just done my calculations, based on g=9.81 m/s^2.

 

I get the wrench dropping at -13.62 m/s after two seconds, i.e. travelling downwards. The distance below the point where it was released is 7.62 m, and in two seconds the helicopter has risen 12 m. Therefore, the separation between the wrench and the helicopter is 19.62 m. Does that look right? :lol:

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Here are some equations that i wrote out for you guys that you might be able to use. (gotta pick the right ones)

 

V=d/t

 

A=Vf-Vi/t

 

Vf=Vi+at

 

V=Vf+Vi/2

 

d=1/2 (Vf+Vi)t

 

d=Vit+1/2at^2

 

Vf^2=Vi^2 +2a

 

I missed physics in high school, but I don't see an equation to plug in the constant for gravity. As far as how to use the answers, I would expect to able to able to round my answer to the nearest thousandth's

so answer a would be the north cooridinates and answer b would be the south cooridinates. the degrees would be the local degrees unless you wish us to do other calculation.

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Please don't take this personally since I know many others like this type of thing and won't agree with me. But...I would avoid a cache like this like the plague.

 

For me its still about the cache location. I would rather do a cache where the placer put his creative mind into the location. Just give me the coordinates and I will get there and gladly share my feeling about what you wanted me to experience.

 

I spent many years in college to get my AS, BS and finally my MS vowing to try and avoid anything remotely like taking an exam or defending a thesis again ...and I especially loathed all those semesters of Physics and Calculus. :P It's not that I mind thinking through something to find an answer, its just puzzle caches like these seem to much like taking a "pop" quiz on Friday afternoon before you can go play on the weekend.

 

Man, I hated that. :o

 

In any event...good luck!

 

Salvelinus

Edited by Salvelinus
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It would take me a while to figure out the math, since at first I would be on a head hunt for whoever the heck unsafely dropped a wrench out of my helicopter!!! :P

 

Then, perhaps, I would get to the math. Like someone else stated, it would not be on my immediate short list to grab a FTF, but as the other caches in my area got completed, I would get to it eventually.

 

Sheesh, and to think some folks think my puzzle caches are "hard." :o

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A wrench falls from a helicopter that is rising steadily at +6.0 m/s. After 2.0s

m/s = Miles Per Second?

 

That's about damnn fast for a helicopter.

 

Did it slide out an open door, was it hanging off a nut, or was it intentionally dropped? Adjustable wrench, open end, box end, or combination?

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In response to your original question, while many would attempt such a cache, I probably wouldn't. It's been more than 30 years since I took Physics. Since I never use them, I've forgotten almost all the formuli. More power to those who enjoy this sort of thing. Fortunately there are plenty of more straightforward caches for those like me who aren't rocket scientists.

 

Æ

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With or without friction from air?

from the OPs first post.

 

ALSO NEGLECT ALL FRICTIONS. (keep it easy)

Looks like I had a case of RTFM.

 

The problem isn't hard when you neglect drag. However not a lot of people will attempt the cache when the math starts getting complex. You limit your pool of finders to the science and engineering types, and the people who do math probelems for fun. A lot of those put down the math books when they get home so you limit your pool of finders even more.

 

Stick a wonderful FTF prize in the cache though and you will generate some interest.

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It's been a bit, so please excuse me if my physics are off.

 

I'm using g=9.82 m/s^2 for simplicity sake.

 

The velocity would be 19.64 m/s in the downward direction

 

And the wrench would be 19.64 m below the point which it was dropped from (also assuming this is the center of mass for the helicopter for simplicity sake).

 

 

I wouldn't mind doing this beforehand, as long as things (such as what you used for gravity) are given.  I'm not sure if it would be widely enjoyed, but I would venture that a fair number of people would enjoy it, as long as they understand the physics behind the question you pose.

 

 

edit: Added response to questions...hehehe

i usually use 9.8m/s/s for gravity just to make calculations quicker...but 9.82 is more exact. However the resulting velocity of the wrench would be a negative number to indicate direction. Positives for up, negatives for down. Your first answer is close but not quite right. Also when finding how far below the helicopter the wrench is, take into consideration how high the helicopter was when the wrench dropped

Why? The height of the helicopter at the time the wrench was dropped has no bearing whatsoever on how far or fast the wrench falls RELATIVE TO THE HELICOPTER, and a negligable impact relative to the earth (edit: see my post below). The wrench will accelerate toward the earth at 9.8m/s/s regardless of whether it was dropped from 200 feet or 10,000 feet (simple physics only, again see below if you want to get technical, which I'm assuming you don't since you're ignoring friction).

 

Since your only question regarding altitude was how far the wrench is below the helicopter, the helicopter's initial altitude is of no concern to us, unless the helpcopter was close enough to the ground so that the wrench would land in under two seconds.

 

As far as your questions go:

 

A wrench falls from a helicopter that is rising steadily at +6.0 m/s. After 2.0s

a. what is the velocity of the wrench?

 

Since velocity is relative, there is not enough information to answer that question. If you are talking relative to the helicopter, then the velocity of the wrench is -19.6m/s. If you are talking relative to the earth, then the wrench's velocity is -13.6m/s.

 

b. how far below the helicopter is the wrench?

 

That one is easy. To answer this one, you don't need to know how fast the helicopter is moving, just it's rate of acceleration. Since you said it's "steadily rising", then the chopper has an acceleration of 0 and the distance traveled by the wrench RELATIVE TO THE CHOPPER is independent of the chopper's speed or altitude. All we need to know is that the wrench had an initial relative velocity of 0, and an acceleration of 9.8m/s/s. Given that d=v(o)t+1/2at^2, the distance between the helicopter and the wrench is 19.6m.

Edited by ParrotRob
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Would the gravitational pull be constant on said wrench at sea level and at higher altitudes (say...hemalayas) where the air is thinner?

Technically, the acceleration due to gravity will be less at a higher altitude than a lower one. This is because of the fact that the gravitational pull between two objects is inversely proportional to the square of the distance between them. Because the distance is measured from the center of gravity (the center of the earth), small distances above the surface have a very limited impact on the acceleration and are generally ignored in simple physics.

 

An object placed above the surface of the earth by an amount equal to the radius of the earth would double the distance, thereby quartering the acceleration.

 

Given that the mean radius of the earth is 6371km, and the acceleration due to gravity is accepted to be 9.82m/s/s at sea level, the acceleration, even at the top of Mount Everest (8850m), would still be 9.79m/s/s.

 

By the way, that part of it has nothing to do with the viscosity of the atmosphere ("thinner" air) and would hold true even if there were no atmosphere at all.

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Is there a free wrench in all this?

No way, the wrench is most definately not "free", having been captured by the gravitational pull of the earth. Interestingly, the earth is also accelerating toward the wrench, as the wrench exerts it's own gravitational pull on the entire planet, believe it or not.

 

In fact, there's no such thing as a free anything, since every object in the universe is attracted to every other object in the universe to some extent.

 

Does that help? :o

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Is there a free wrench in all this?

No way, the wrench is most definately not "free", having been captured by the gravitational pull of the earth. Interestingly, the earth is also accelerating toward the wrench, as the wrench exerts it's own gravitational pull on the entire planet, believe it or not.

 

In fact, there's no such thing as a free anything, since every object in the universe is attracted to every other object in the universe to some extent.

 

Does that help? :P

Are you attracted to me? :o

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Is there a free wrench in all this?

No way, the wrench is most definately not "free", having been captured by the gravitational pull of the earth. Interestingly, the earth is also accelerating toward the wrench, as the wrench exerts it's own gravitational pull on the entire planet, believe it or not.

 

In fact, there's no such thing as a free anything, since every object in the universe is attracted to every other object in the universe to some extent.

 

Does that help? :P

Are you attracted to me? ;)

Ok I was getting a head ache from the math but now I've got one from laughing :o

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I've just done my calculations, based on g=9.81 m/s^2.

 

I get the wrench dropping at -13.62 m/s after two seconds, i.e. travelling downwards. The distance below the point where it was released is 7.62 m, and in two seconds the helicopter has risen 12 m. Therefore, the separation between the wrench and the helicopter is 19.62 m. Does that look right? :D

Nice job on the calculations. they are exactly right....

 

And to everyone else thanks for the awesome responses!

 

Anyways...on to answering some of your questions, my cache when it is placed will be on top of a small 1.2 mile trail which overlooks the local lake.

 

-m/s is not miles per second, its meters per second.

 

-Yes, my last formula was incorrect i added the d at the end

 

Now so everyone can understand it better, neglegt all frictions. This mens when you drop a feather or a wrench they will fall at the same 9.8m/s^2 because there is no air friction, so it doesnt matter if im on everest or at sea level.

 

Now to explain all that Vi,Vf stuff....lol

 

Vi=initial velocity

d=distance

a=acceleration

Vf=final velocity

t=time

 

ect....

 

Thanks everyone for your responses, at the moment im working on thinking up another problem if you guys want to solve it or discuss it, or whatever go for it!

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