A puzzle which has occupied me for many years -
You have to go from A to B a distance x. If it is raining, will you get less wet if you run?
I don’t have an answer, but it is an interesting question.
I’ve wondered about the rain problem. Assume the same amount of rainfall throughout the journey. Approximate your head as a circle (or any other shape) perpendicular to the rain, which is approximated to fall vertically. Approximate the front of your body as a planar surface parallel to the rainfall. Various elementary school classes have demonstrated that your head gets less wet if you run, and I think that’s intuitively obvious. How many more raindrops the front of you encounters is going to be a function of the velocity of the rainfall and how fast you run. The relative size of the two planar surfaces comes into it as well.That’s as far as I’ve been able to get.
This is a classic and much argued over problem. The types of analysis most of us would head for are formalised here:
http://www.people.hbs.edu/dbell/walk%20or%20run%20in%20the%20rain.pdf
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You are right that relative velocity is the correct measure of damage (basically the kinetic energy at impact).
You still have some head scratching to do to get to the right answer. All the information you need is in the question and the answer will likely surprise you ( as it did me ).
I don’t think the velocity of the rain matters. I think it is the related to the volume of space you travel through in unit time and the rain flux. I see IanG has provided one analysis.
Assuming the tether is fixed and can’t slip around the fence the tether should be a bit longer than the radius - if it is exactly the radius it will inscribe an area less than the semicircle. I’ve tried to work out how much longer but don’t see a way to calculate it without some ugly and therefore tedious maths which is no fun. Hoping for some inspiration on the drive home.
Ah! You anticipated my next teaser… 
This is a (badly) drawn fulcrum of a square base right pyramid.
The base is a 4 x 4 square and the top is a 2 x 2 square.
The perpendicular height is 6.
What is the volume of the fulcrum ?
(Sorry Ian, I will post the (really hard) Goat in a Field teaser shortly. But you have a head start, since you have visualised the question)
I think Ian’s link provides an excellent answer. Basically … RUN
If the up-going elevator is fairly close to the top of the shaft, the brick might not have reached terminal velocity and thus cause little damage. If the down-going elevator is near the bottom of the shaft, the brick at terminal velocity could cause considerable damage.
I presume we are dealing with a brick at terminal velocity colliding with the elevator at full speed, regardless of whereabouts they are in the shaft ? I’m visualizing two such identical bricks, simultaneously hitting a full-speed up-going elevator and a full-speed down-going elevator as the elevators pass each other.
Another easy one to provide “hope” while we contemplate the elevators and bricks. The other easy one is the fulcrum.
One amount added to a quarter of that amount becomes 15. What is the amount ?
Don
Assuming the tether is fixed and can’t slip around the fence the tether should be a bit longer than the radius - if it is exactly the radius it will inscribe an area less than the semicircle. I’ve tried to work out how much longer but don’t see a way to calculate it without some ugly and therefore tedious maths which is no fun. Hoping for some inspiration on the drive home.
Well, the tether IS FIXED and can’t slip and there is a way to calculate it, to any number of significant figures, however, it does involve ELEGANT maths to start, followed by some very tedious maths to get a solution. Nowadays, Excel can help to reduce the tedious part.
So just to remind others, the grass field is circular, let’s say radius 100m. The goat is tethered to a fixed point on the perimeter. The tether (rope) is just long enough to allow the goat to eat half the grass. How long is the rope. Ignore the distance between end of rope and mouth of goat !
As kenc would say “enjoy”
12 unless I’m missing a subtlety.
Hi Eoink,
There was nothing subtle about it and you’re not missing anything.
12 it is and well done ! (even if it was “too” easy !)
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When I did my degree there were 2 papers in the final year called the Comprehensives, which covered fairly basic Physics, A’Level and 1st year degree, the idea being to set “real world” type problems and see if we could actually apply the stuff we’d learnt, rather than just regurgitate it. Each paper was 10% of your whole degree marks. I still remember doing some past papers and working this through, I suspect I got to a less sophisticated version of the paper Ian linked to.
I don’t know if I ever said this on the old board Don, but thanks a lot. I enjoy the occasional stretching of the little grey cells, and I really appreciate the effort you put in to give us these challenges.
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I should have added you can ignore the effect of air resistance so the brick never reaches terminal velocity but just accelerates due to gravity.
Thank you Eoink.
I enjoy working through them before posting.
Ah !
And the elevators move at a constant speed, or if starting from rest, they accelerate far more slowly than the falling brick - well, that’s the idea behind an elevator, I hope
Thus the down-going elevator provides room for the brick to accelerate even faster, whilst the up-going elevator reduces the room for the brick to accelerate etc.
Yes the elevators move at constant speed. I think you’ve got the picture now. If it helps you can think of the elevator being a height h below the brick when it begins to fall, but the value of h is not important.