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archery release arrow
I need help deciding if a pair of sneakers or a series of archery?
The pair of Nike SB are Blue Lobster 10, which were released very limited. They cost $ 200. I have about 10 brands in the vicinity of the new Air Jordan $ 150 all. I will be the final year of secondary school next September. So the last year before college as a college, I'll be more focused on education in fashion. The second element is a Martin Takedown Recurve Bow Jaguard kit comes with basic needs such as arm guards, arrow, Stringer, etc. and costs $ 200. I need help deciding what to buy and why you buy it?
Well, if you already have about 10 brand new shoes I would go with the archery world, and it looks really cool worthwhile if you use a lot. $ 200 shoes are a bit too lol
archery release arrow
I need help deciding if a pair of sneakers or a series of archery?
The pair of Nike SB are Blue Lobster 10, which were released very limited. They cost $ 200. I have about 10 brands in the vicinity of the new Air Jordan $ 150 all. I will be the final year of secondary school next September. So the last year before college as a college, I'll be more focused on education in fashion. The second element is a Martin Takedown Recurve Bow Jaguard kit comes with basic needs such as arm guards, arrow, Stringer, etc. and costs $ 200. I need help deciding what to buy and why you buy it?
Well, if you already have about 10 brand new shoes I would go with the archery world, and it looks really cool worthwhile if you use a lot. $ 200 shoes are a bit too lol
archery release arrow
Math HW help? Archery problem?
I use to know how to do this but because of summer I completely forgot, I just know it looks familiar.
Archery Problem: An archer climbs a tree near the edge of a cliff, then shoots an arrow high into the air. The arroes goes up, then comes back down, going over the cliff and landing in the valley, 30 m below the top of the cliff. The arrows height, y, in meters above the top of the cliff depends on the time, x, in seconds, since the archer released it.
the graph show the point (0,5) as where you where when you launched the arrow and after 2 seconds is the highest point (2,25) the arrow goes up till it heads down again. Because your on top of a hill the arrow falls off the cliff and the lowest on the y-axis when it hits the ground is -30 but the x point is a number between 5 seconds and 6 seconds.
A) What was the approximate height of the arrow at 1 second? at 5 seconds?
I forgot how to do this sort of problem.
use the time and distance formula for an object undergoing uniform acceleration.
S(t) = So + Vot + At²/2
Where:
So = distance of the object from your zero datum at time t=0. Units of length.
Vo = velocity of the object at time t = 0. V is positive when it is in the direction of increasing S and negative when in the direction of decreasing S. Units of length per unit time
A = constant acceleration. Positive in the direction of increasing S and negative in the direction of decreasing S. Units of length per the square of unit time.
Since you are dealing with metric measurements in the Earth’s gravitational field, the numerical value of acceleration will be 9.8 m/s².
The points quoted implies that the datum S = 0 is at the top of the cliff and the tree from which the arrow is launched is 5 meters above the top of the cliff. That determines S(0) = 5. And since gravitational acceleration is directed downward, in the direction of decreasing S, we take the acceleration to be -9.8 m/s. Using what we know so far, the equation of motion is:
S(t) = 5 + Vot – 4.9t²
Since the statement is made that the highest point on the arrow’s path is S(2) = 25, we can use that to determine Vo.
S(2) = 25 = 5 + 2Vo – 4.8(2)²
20 = 2Vo – 19.2
39.2 = 2Vo
Vo = 19.6 m/s
Our equation of motion for the arrow’s flight is now complete:
S(t) = 5 + 19.6t – 4.9t²
Approximate height of the arrow at 1 second:
S(1) = 5 +19.6 – 4.9 = 19.7 meters
(19.7 meters above the edge of the cliff)
Approximate height of the arrow at 5 seconds:
S(5) = 5 + 19.6*5 – 4.9*25 = -19.5 meters
(19.5 meters below the edge of the cliff)
How long after release does the arrow strike the ground at the bottom of the cliff?
S(t) = -30 = 5 + 19.6t – 4.9t²
4.9t² – 19.6t – 35 = 0
t = (1/9.8)[19.6 ± √(19.6² + 4*4.9*35)
t = (19.6 ± 32.713)/9.8 = 5.338 seconds
(we can discard the negative value since we are only interested in the time after the arrow is released)
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