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I have been trying to solve this mystery geocache called "Clas of 43" on geocaching.com and would appreciate any help. The part I am stuck on is calculating the distance the cannon ball travels. If I can get that right, I can solve the rest of the mystery using the great circle calculator link on the cache page. Here is a link to the cache:
http://www.geocaching.com/seek/cache_details.aspx?guid=55b5bf3a-1a0f-482f-93dc-2341dede5ebb

2006-08-17 00:55:10 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Got it figured out. The way that I figured it out was calculating the amount of time that the cannonball is aloft from v = gt, noting the parabolic course will make it twice the time from initial velocity in the vertical direction (vy = v sin 31.5 deg) to zero velocity. I got about 19.623 s. Then I calculated the distance traveled in the horizontal direction using x = v * t * cos 31.5 deg to get 3079 metres.

There's a few other caches that use this type of problem. One near where my brother lives is this one (http://www.geocaching.com/seek/cache_details.aspx?wp=GCGTRZ). Just thought that I bring that up for kicks.

I wish you good luck on your hunt.

2006-08-17 13:48:53 · answer #1 · answered by Ѕємι~Мαđ ŠçїєŋŧιѕТ 6 · 0 0

I also am a geocacher so I know what you are talking about.

Google the word Kinematics. You will find a set of equations that are used to make ballistic calculations. And while I thank you for the 2 points here, I hope you will get your smiley face at GC.

Beware of geomuggles and happy hunting.

2006-08-17 01:03:00 · answer #2 · answered by sparc77 7 · 0 0

Its a remember of breaking down the vertical and horizontal vectors of the launch. the thank you to do it is in the link below. The horizontal displacement is around 3078.89 meters in the path indicated on the cache training (281 stages i think of) I used 9.8 meters consistent with 2nd for the gravitational acceleration. the internet internet site makes use of 10.

2016-10-02 04:50:39 · answer #3 · answered by boscia 4 · 0 0

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