How far could an arrow go?

[hutchies]

All depends on Trajectory. PM AJ he does a buch of balistics stuff and I bet he could guestimate it. My guess with the right angle. 1/2 mile or so.

[Master_Chief]

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/...ic_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.

[joebobhunter4]

holy crap are u a math wizard or what?!?!?!?!?!?

[Griz]

[ QUOTE ]
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/...ic_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.

[/ QUOTE ]



Yep! That should do it...
Say, can we use beer cans as an example instead. I think it might help me understand this just a little bit better...

[Rhino]

[ QUOTE ]
[ QUOTE ]
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/...ic_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.

[/ QUOTE ]



Yep! That should do it...
Say, can we use beer cans as an example instead. I think it might help me understand this just a little bit better...

[/ QUOTE ]

Need a little ciphering fluid there eh Dale.

[buckee]

LMBOROF
Lost me ...LOL ..but I believe ya

[tyshe17]

If yall were real archers you could calculate that in your head and know the formula by heart!!!

[ThethirdI]

Stick out your thumb as if you were trying to hitch a ride and point it at the horizon. Then say outloud in a stern voice to yourself "About that far I reckon!" That's the formula I would use. LOL

[Griz]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
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/...ic_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.

[/ QUOTE ]



Yep! That should do it...
Say, can we use beer cans as an example instead. I think it might help me understand this just a little bit better...

[/ QUOTE ]

Need a little ciphering fluid there eh Dale.

[/ QUOTE ]

LMBO...
Yeah even a few corks out of the jug might do it to...
Geez I was in school for 25 years and they never had nothing like this in the 3rd grade...

[johnf]

The longest shot I've heard of was 320 yards. If you use the formula for getting the circumgerence of a Circle then divide it by 2 you get 502.4 yards or around 1507 feet of total movement. There are 1509 feet in a mile, so then the arrow would fly just two feet under a full mile, give or take a little.




I can't belive I just took the time to figure that out.

[huntinsonovagun]

A maximum of 236 ft??? That is 78.667 yards.....I practice that far practically....my bow will shoot farther than that.....most bows will shoot farther than that.

Now, that formula is right if you are shooting horizontally, but if you shoot it at say a 45* angle, you can shoot A LOT farther.....trust me....

I have always wanted to fling an arrow, but they are too expensive to do that.

I'm sure if you find the right person, they could tell you.

[AJ]

[ QUOTE ]
The longest shot I've heard of was 320 yards. If you use the formula for getting the circumgerence of a Circle then divide it by 2 you get 502.4 yards or around 1507 feet of total movement. There are 1509 feet in a mile

[/ QUOTE ]

You have some huge feet then. How about 5280 feet in a mile or 1760 yards in a mile.

[maddhunter]

Me and a buddy of mine shot our bows out into his pasture at about a 45 degree angle and I range it at 685 yards.

[huntinsonovagun]

OK, so at 685 yards, which is 2055ft, that would mean this arrow went just under .4 of a mile. More than likely, Dwin (madhunter) and his buddy didn't have their bows at the optimum angle for distance, so it's possible the arrow could have gone a little farther...... You plan on becoming a SWAT sniper with a bow or something?!?!

[gfourhunter]

just shoot the bow
at a deer 30 yards ( 90 feet ) and you will be ok.