Master_Chief

Re: red light cam (funny) That's hilarious. I bet they laugh at that one all the way to the subpeona.

Re: put one up ROFL. Thanks guys.

Re: First Frost!!!! Great for you. Still looking at green, green here. Hoping it will at least smell like fall soon.

Re: Big Buck... Looks like a nice one.

Re: Couple Bear pics Great pics. Good luck.

Re: Got bored Very nice.

Re: My good deed for the day Nice job 10.

Re: 7 Hours till I go!! Pray please!!!!! Be careful and safe. Prayers for a safe return.

Re: my ground blind Nice job. Good luck.

Re: The place i call camp..... Looks great.

Re: some bad news Bummer. I was hoping to see you with him under your stand.

Re: First Deer Down and in the Cooler!!! Great job. How was the sweet meat?

Re: Here is my 2005 NC BUCK Congrats on the nice deer.

Re: Got a new job today! Congrats on the new job.

Re: New Puppies!!!(pics) Awesome pics. Congrats on the pups.

Re: My Date with Mommy. Nice one. Thanks for serving.

Re: why are deers necks so big in the rut? You guys crack me up.

Re: Been out for a bit Sorry for your loss.

Re: More Fun Than I Can Stand!!!! Hope you didn't bend the gate.

Re: hunt in any state? Iowa for sure.

Re: i just can\'t wait!!!!!!! Save me a piece. I don't get out in Va. but will be heading to Pa for the last week of archery. Cousin has 200 acres in Clarion County. Good luck and post your pics.

Re: Monday morning doe Good job

Re: not a wasted saturday Good news for you. Good luck.

Re: Scrape Info Thanks for the read.

Re: How far could an arrow go? 1/2 mile seems a little far. That is 880 yards. I would be surprised if it would go 440 yards. Check this formula out for how far it will go if you miss a target. An arrow is shot from a point 20 feet above the ground with an initial horizontal Velocity of 45 feet per second and an initial vertical velocity of 80 feet per second. The parametric equations that describe the arrows path are: x(t) =45t y(t) = 20+80t -16squared The arrow is aimed at a target that is horizontal distance of 90 feet from the point where the arrow is shot. If the arrow hits the target, how high is the target? If the arrow doesnt hit the target, how far does the arrow travel before it hits the ground? Solution 28305 x(t) = 45t y(t) = 20 + 80t – 16(t^2) a) Arrow hits the target: x(t) = 90 45t = 90 45t / 45 = 90 / 45 t = 2 Find the height of the target when t = 2: y(2) = 20 + 80(2) – 16(2^2) = 20 + 160 – 64 = 116 The height of the target is 116 ft. b) Find the time when y(t) = 0, and t > 0: y(t) = 0 20 + 80t – 16(t^2) = 0 [20 + 80t – 16(t^2)] / 4 = 0 / 4 5 + 20t – 4(t^2) = 0 (-4)(t^2) + 20t + 5 = 0 Apply quadratic formula: http://www.gomath.com/htdocs/lesson/quadratic_lesson3.htm t = {-20 – sqrt[20^2 – 4(-4)(5)]} / [2(-4)] = {-20 – 4[sqrt(30)]} / -8 = (5/2) + (1/2)[sqrt(30)] Find the horizontal distance traveled by the arrow: x(5.24) = 45(5.24) = approximately 236 The arrow travels 236 ft before it hits the ground.