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.