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GAG23 – As Easy as Falling Off a Cliff

A cache by MAPSIT Send Message to Owner Message this owner
Hidden : 10/09/2015
Difficulty:
3 out of 5
Terrain:
5 out of 5

Size: Size: small (small)

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Geocache Description:

A boat accessible mystery cache placed for the GAG23 event


The cache is not at the posted coordinates, though it is within a kilometer of them. If you want to go there, that’s your business, but I have no evidence that the fishing there is particularly better than at other spots in the Ottawa River. In keeping with the spirit of the GAG event, please do not attempt to look for the cache before 18:00 on Friday, October 16, 2015, by which time I promise to have removed all of the Acme Corporation devices currently protecting the cache container.

 

The cache, a small camo’d sandwich box, is at N45 24.ABC, W075 45.DEF, but to work out the values for A through F, you’ll need to (a) first answer some questions about falling off a cliff, (b) use those answers to obtain the correct single digit values for variables G through L, and then (c) do some simple math. By way of a check to help keep you from wandering around looking in the wrong spot, the digital root of the sum (A + B + C + D + E + F) is 6.

 

Followers of the Wile E. Coyote versus the Road Runner cartoon will undoubtedly have noted that Wile E., in pursuing the Road Runner, regularly sees his schemes fail, often with disastrous consequences. Wile E. gets caught in explosions, or crushed by falling weights, or he smashes into one rock wall or steel plate after another. However, some of the most memorable outcomes involve him falling off a cliff, typically exhibiting a confused or agonized expression as he presumably anticipates just what is going to happen to him a few seconds later. To find this cache, you’ll need to do some relatively simple, non-academic-stream-high-school math involving falling off a cliff. You might also want to consider taking advantage of the Pythagorean Theorem. And if your free-body physics skills have, like mine, grown ever so slightly rusty, you might want to consult –

http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall.

 

First let’s set up the scenario. Wile E. is running across the top of a large flat/horizontal mesa, situated 0.4971 furlongs above a large flat plain down below. The edge of the mesa in the vicinity of the relevant area is, for purposes of this exercise, a horizontal straight line. Wile E., having been tricked by the Road Runner, is heading directly (perpendicularly) toward that edge, as if, seen from above, he is coming up the stem of a “T” toward it. He is running at a speed of 31.0686 miles per hour. (Wile E.’s maximum speed is only about half that of a cheetah, and of course slower than that of the Road Runner, lest the cartoons get boring very quickly and not be rated kid-friendly.) In this cartoon universe, gravitational acceleration is 10 m/s/s (downward), a bit higher than in the real world, because as you well know, in cartoons, things have to happen faster than in real life, lest viewers with short attention spans get bored. Also because this is a cartoon universe, feel free to safely ignore matters of air resistance, the variation of the gravitational force as a function of the distance from the center of the earth, both the earth’s curvature and its rotation, and any and all relativistic effects.

 

1)Calculate the time that elapses, from the instant at which Wile E. passes the edge of the cliff, to his impact on the plain below (in seconds). The URL may prove handy. Let G be half of this time, rounded to the nearest whole second.
 

2)Compute the distance (to the nearest whole meter) from the point at which Wile E. leaves the cliff (up top), to his point of impact on the plain (down below). Don’t worry about the length of the curved arc that he follows, you just want the straight-line (diagonal) distance between his up at the top starting point and  his down at the bottom ending point. Let H be the final digit of this rounded number of meters.
 

3)What is the speed with which Wile E. hits the ground (in yards per minute, to the nearest whole yard per minute)? (Don’t forget to include both the horizontal and vertical components when calculating this speed.) Let I be the final digit of this rounded yards per minute speed.
 

4)I neglected to mention that the cliff is not straight up and down, though it’s steep enough that Wile E. falls all the way to the plain below, rather than impacting on the cliff slope. Indeed, erosion has somehow worn the entire cliff to a steep, uniform angle of 30 degrees, measured from the vertical. Knowing this, calculate the distance along the plain, from the bottom of the cliff slope (measured along the line underneath Wile E.’s trajectory,) on out to his point of impact. Express your result in fathoms, rounded nearest whole fathom. Let J be the final digit of that rounded value.
 

5)The very next day, tricked by the Road Runner yet again, Wile E. (miraculously recovered from the previous day’s experience, and as speedy as before) leaves the same cliff at exactly the same spot, with exactly the same speed, and once again horizontally with respect to both the top of the mesa and the plain below. But this time, rather than heading straight out, he leaves at an angle of 15 degrees off of his previous path, i.e., the angle between his path and the edge of the cliff (as seen from above) is 75 degrees ( or 105 degrees) rather than 90 degrees. Assume that, once he recovers from the impact, he crawls (probably rather slowly and painfully) to the closest point along the bottom of the cliff slope to begin his climb back to the top. Figure out the length of that crawl (in meters, rounded to the nearest whole meter). Let K be thrice that rounded value.
 

6)(OK, OK, having seen Wile E. get all his hair blown off by explosions too many times, I just couldn’t get that image out of my mind, and felt compelled to include something related to it in the puzzle for this cache. No cliff in this part. So sue me.) Suppose that Wile E. has once again lost all his hair in an explosion. So just how many hairs will he need to regrow? Curious minds want to know! To help you in your calculations, I can supply you with the following data. Assume that, (a) consistent with scientific studies, Wile E. has about 300 hairs per square centimeter on his furred portions, (b) Wile E. is (very conveniently) topologically equivalent to a sphere with a one meter diameter (i.e., assume a spherical coyote one meter across), and (c) 5% of his surface area is not covered in hair, being devoted to important, inherently non-hairy, body parts such as his nose, his eyes, and the pads on his feet. Calculate the number of millions of hairs (rounded, since the 300 hairs per square centimeter figure is only an approximation, to the nearest million) and let L be the digital root of that rounded number (of millions of hairs).

 

And now, using the values of digits G through L that you’ve generated, calculate the necessary digits A through F from the following formulae.

 

A = I – 6*(H – K – J)

B = (G + H +J + K)/3

C = L – H – (G/J)

D = K/J

E = L- G - J

F = G + I + J

 

Finally, get out there and find that cache. And (please) use a great deal more caution and common sense than we typically see from Hungrii Flea-baggius.

Additional Hints (No hints available.)



 

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