Mystery Geocache - Where will it land??

Question

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

Answer 1

The formula for finding the range is:

v^2 sin [ 2 (theta) ]/g

v is the velocity (184.025 m/sec)
theta is the angle (31.5 deg) and 2 theta is 63 degrees.
g is gravitational acceleration (9.8 m/sec^2)

That's about 3079 meters, or nearly 2 miles.

Answer 2

I can confirm the calculations already made. 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.

Answer 3

Its a matter of breaking down the vertical and horizontal vectors of the launch. HOW to do it is in the link below.

The horizontal displacement is around 3078.89 meters in the direction indicated on the cache instructions (281 degrees i think)

I used 9.8 meters per second for the gravitational acceleration. The website uses 10.

Answer 4

No Idea - sorry !

May be you can do some research on the www it self.