Introduction: Welcome to the artillery range
officer. You are currently located on top of a small hill some
20m above the river flat below. Your task is to
target a small strategic point just beyond the river. Rather than
provide you with grid coordinates you will be provided with bearing
and range data and you are to locate the target from that. Further,
to ensure your full understanding of the maths and physics behind
this sort of activity you will be required to find the bearing and
range data by solving some simple problems.
Bearing: The bearing can be found if you take a
100g mass on the end of a 0.80m
length of string and spin it clockwise (that is it passes through N
then E then S then W) in a horizontal circle such that its
instantaneous velocity is 4.3ms-1. If
the mass passes through 0° (true north) at time, t=0, it will pass
through the desired bearing at exactly 23.0 sec.
It will, of course, pass through this bearing a number of times in
23 seconds but it will be at the correct bearing at 23.0 sec.
Range: The launching system you are using fires
its projectile at exactly 60.50ms-1. The launcher is
angled at 37.5° to the horizontal. This will give you the correct
range to the target.
Once you have the target located you may fire when ready.
Assumptions and simplifications: The task will
require the use of VCE year 12 Physics. You will need to apply an
analysis of uniform circular motion and projectile motion,
specifically bi-level projection from an elevated position. Further
you will also need to apply year 9 trigonometry. This is NOT an
exercise in the finer points of cartography or the vagaries of
Mercator projection. For the sake of this exercise you may consider
one minute of both latitude and longitude to be 1852m exactly. I
realise that this only holds true on the equator and for those of
you planning to use a range and bearing facility in your GPS I
apologise but the puzzle is about physics, not geography. Also, in
keeping with the year 12 curriculum, you may assume acceleration
due to gravity to be 10ms-2 (you should know that it is
in reality 9.8ms-2) and you may ignore air
resistance.
Current GZ info: The cache is a 2L systema
container painted black. It has been placed in a location obvious
to cachers and lies about 5m off the trail. Beware of muggles -
this track is heavily used if the weather is good.
Getting help: If this puzzle is really beyond
you maths and physics ability to solve then you can make use of a
worked solution I have uploaded to the Victorian Government's
Education Resource website: Fuse. Point a browser at
fuse.education.vic.gov.au and enter GJ7KQD in the Learning Resource
ID box on the right hand side. You don't have to be a teacher or a
student to use this resource. I would ask that if you do use this
resource that you let me know either by email or in a log and let
me know how easy you found it to follow.
Acknowledgments: Thanks to my brother KJ whose
skills as a town strategic planner and urban designer were much
appreciated in finding this location and for putting me on to
land.vic.gov.au without which I could not have designed this
puzzle.
Thanks also to Trailrunning for testing my calculations and
verifying that they do, indeed, lead to the cache location.
Trailrunning has also provided a FTF pathtag as a prize.
Thank-you.
