Posted: 5/19/2001 5:22:15 PM EDT
Hypothetical:
You want to shoot a monkey in a tree.
The monkey is 30' above the ground. You are 100 yards away on the ground.
Where do you aim? One stipulation: the monkey drops the exact instance you pull the trigger.
A) Above the monkey
B) At the monkey
C) Below the monkey



below



Aim dead on the monkey.



You miss the monkey since you had no idea it was going to drop as soon as you pulled the trigger.



Originally Posted By Righteous Kill:
Aim dead on the monkey. View Quote 


dead on, but the monkey will fall when the bullet hits him, not when i pull the trigger.



This is assuming the rifle is not sighted for 100 yards. You are aiming the barrel with no sights.
The bullet and the monkey drop at the same rate. Not factoring in air resistance.



You're bullet is travelling at better then 3000 feet per second. The monkey is falling 32 ft/sec2. At 100 yards the shot is thought the monkey before he knows it. Shots made up or down always hit higher than the point of aim.


Celebrating the Second Amendment One Fine Firearm at a Time

Originally Posted By SILVER SURFER:
dead on, but the monkey will fall when the bullet hits him, not when i pull the trigger. View Quote 


ok I think I can do this one, aim 3.11 inches low a 3600fps bullet would take .083 seconds to clear 100 yards lets say .09 to account for locktime so by the time the bullet is there the monkey is falling at 2.88 feet per second, at .09 seconds, he's fallen 3.11 inches. of course your actually shooting a bit more than 100yd since he is 30 feet up but damn now that I thought of that I'm not refiguring and also the bullet has slowed down a little bit by the time it gets 100 yds out..



Simple 
81 mm HE  nothing left but Monkey Shit
[:D]


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Escape from NY the Sequel <font color=red><font size=3>RENDELL MUST GO!</font id=s3></font id=red> "ATF  Proudly Mocking the Constitution since 1968" 
Originally Posted By Janus:
The bullet and the monkey drop at the same rate. Not factoring in air resistance. View Quote 


Originally Posted By erickm:
2.88 feet per second, at .09 seconds, he's fallen 3.11 inches. View Quote 


At the monkey of course.
Why? Because the monkeys acceleration (from gravity) is the same as the bullets, regardless of the bullets speed. 9.8M per second, per second.


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Originally Posted By Tex78:
At the monkey of course.
Why? Because the monkeys acceleration (from gravity) is the same as the bullets, regardless of the bullets speed. 9.8M per second, per second. View Quote 


Everyone who said at the monkey would kill it but not hit it where they wanted to. Duh the monkey is 30' up in a tree. therefor your bullet must rise close to thirty feet before it hits the monkey. Now a 100 yards it 300 feet therefor the distance the bullet must travel is = to (30^2 + 300^2)^(1/2) which is. 301.5 feet. The angle at which the bullet is going relative to the gound is inverse tangent of (30/300) which is 5.7 degrees. Now since gravity only works staight down then the monkey will accelerate at 9.8 m/s/s or 32f/s/s, but the bullet will not. depending on the velocity of it it will only accelerate differently. lets say the bullet is going 3000f/s. so it velocity in the x direction would be 3000*cos(5.7) or about 2985f/s which is 15f/s less than the magnitude. Now it if the bullet has to go 300f in the x direction to hit its target it will take 300/2985 or .1005 seconds. In that time the monkey will have fallen a vertical distance of .5(32)(.1005)^2 or .1616f. But the bullet on the other hand will have fallen (2985)(.1005)(.5)(32)(.1005)^2 or .1691f.
Now what does all this mean you ask? Well it means tht if you aimed right at the monkey you would have missed by .1691.1616 or .0075 of and inch. Jesus Christ and you guys call yourselves monkey hunters.



This of course not factoring in air resistance which just feaking complicates things with integration since theair resisance changes as the velocity does, but if we were onw the moon this is what would happen.



For those who said right at the monkey Back to vector math with you!
The bullet is not falling it's still rising. The acceleration will not overcome the upward velocity at 100yards. All it will do is slow down the bullet, very little at that.
IIRC you always aim low when shooting at a target higher and lower than you are. If the monkey drops it just means you have to aim a bit, a very little bit, lower.
Enigma, I think you forgot to include the vertical componant of the bullet's path. (As you pointed out it's being fired up, so it has a horizontal componant (x) and a vertical componant (y). I don't think gravity's acceleration would overcome the vertical componant at such a sort distance.) I haven't done the math, but I'm almost 100% sure the bullet wouldn't start falling at 100 yards.


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Aegis Ballistic Missile Defense. It IS rocket science! 
Originally Posted By erickm:
ok I'm correcting myself here I'm wrong, at .09 seconds he HAS accelerated to 2.88fps, 3.11 inches is .09 second fall at 2.8fps but he didn't start the fall at 2.8 he had to work up to 2.8 in that time..how do you figure how far an object drops in .09 seconds, I know what it's velocity is at that time but how much distance did it cover getting to that velocity? is it one of those buttons on my calculator that I don't know what it does? View Quote 


Even the most prudent antigunner would agree this is one of those situations where the hunter would best be served with Full Auto. A quick three second burst would get that darned monkey.



Try thinking about it this way. Say you have a rifle sighted in at very close range, with the barrel parallel to the ground. Then you aim at a monkey far away, the monkey will drop the same amount the bullet does in the time it takes the bullet to get there, and you will hit the monkey where you aimed.
A 3000fps bullet will cover 100yds in 0.1 seconds, in which time it will drop about 2". If you sight in at this distance, you will have already compensated for this drop, so you will hit the falling monkey about 2" high. So if you aimed for center of mass, you would hit the monkey. If you were trying to show off and get it in the forehead, you might miss.



I had to prove this problem on a General Physics test several years ago. Our proffesor even did a demo with a blow gun. As the prjectile, a steel ball left the end of the barrel it released the monkey, a steel can held by an electromagnet. To hit the monkey you aim directly at it.


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Unless the shooter is trying to be too cute, ie a 5 shot .3" group? It really doesn't matter much, except to show why a sniper never headhunts.... [:D]
OTOH??
If the monkey drops faster and the bullet moves slower than mathematically possible?
Maybe the bullet is .5" high on the heart. [:P]



Who is this monkey you are all talking about?
I always want to identify my target before I shoot it! [(:)]


No apology is necessary among friends
A man should never get so comfortable, eat so well, or get so civilized, that he loses his edge for survival 
OK all you supposed mathematicians and physicists using assumptions and almost correct equations. I used actual math and figured out the correct answer (assuming a vacuum).
You aim dead on the monkey and you hit dead on the monkey.
The position of a falling object is equal to the acceleration multiplied by the time squared plus the vertical velocity multiplied by the time plus the starting height. at^2 + vt + h.
thats 32ft/s^2 * t^2 + 0*t + 30ft for the monkey and 32ft/s^2 * t^2 + v*t + 0 for the bullet.
Since you calculate the verticle velocity using the same angles you used to calculate the total distance and time to target, the second component of the equation above for the bullet always comes out to 30ft unless you did bad math. That means the equations are equal since the acceleration and time for both objects is the same. In other words, the small amount of bullet drop off of straight line is exactly equal to the monkey drop.



Use a shotgun and point it right at the monkey.



Hey Ed, [:D]
Some serious math here, but....
I would be more interested in learning this...
As the monkey drops and the bullet hits him in the chest?
How far will his downwards motion be changed (edit: in inches) as he is thrust sideways from the force of the impact? [;)]



HDR
That depends on the mass of the monkey and of the bullet.
If you assume that the bullet doesn't exit the monkey and you know the mass and velocity of the bullet before it enters the beast you can calculate the engery or momentum (mass x velocity) of the bullet before in hit him. You can then calculate the monmentum of the monkey and bullet after inpacts. If you assume that the impact is inelastic, no energy is lost during the impact due to deformation of the monkey, and that the bullet hits the monkey in the center of mass, no energy lost to rotation of the monkey, the momentum before and after shouls be the same. So the velocity of the monkey after impact would be
V(m+b) = M(b) x V(b) / M(b+m)
Where M(b) = mass of the bullet
V(b) = velocity of the bullet before impact
M(b+m) = mass of the monkey and the bullet
V(m+b) = velocity of the monkey and the bullet
If you know the elevation of the monkey when the bullet impacts you can determine the time before the monkey hits the ground using the equation
D=1/2 A T^2 + (Vv x T)
Where
D = Elevation in feet
A = Acceleration due to gravity
T = Time
Vv = Vert. velocity of the monkey when the bullet hit him
Once the time before the monkey hits the ground after being struck by the bullet is determined you can use that to determine how far the monkey will move horizontaly before striking the ground using the following equation
X = V x T
Where
X = Hort distance
V = Velocity of the monkey after being struck by the bullet
T = Time before the money strikes the ground
If the bullet exsist the monkey you need to consider the amount of engery transfered as the bullet traveled through his/her/its body. To do this you will need to determine the velocity of the bullet before and after passing through the monkey as well as any change in mass of the bullet and or the monkey from the impact.


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