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You Math Smart? Problem -- Two Guns

THeMaCHinE

THeMaCHinE

New member
A guy is holding two guns. One pointed at each side of his head. Both guns fire bullets that travel at a velocity of 672 fpm (204 m/min.). Assume friction isn't a variable. The man is facing south. There is a southwestern wind of 28 mph (45 kmph). The left-hand gun muzzle is 1.5" (3.8 cm) from his left temple. The right-hand gun muzzle is 6.25" from his right temple. Assuming both guns were fired at exactly the same moment, which bullet would hit his head first? Or would they both hit at the same time? Support your answer with the mathmatical equation you used to work it out.

Let's see it smart guys/girls...
 
1) if friction doesn't matter, then the wind doesn't need to be taken into account.

2) you didn't tell use the length of the muzzle. one could be a pistol and one a shotgun. the hotgun could be flush against his head and the pistol could be two inches away, but its bullet is still technically closer...

need more info
 
The one from the left hand gun

1.5" < 6.25"

assuming that the both bullets travel at the same rate of speed..... all other info is just B.S.
 
if the guns are the exact same guns, then yeah. the left.
 
WHEN THE BULLET ENTERS HIS HEAD AND PROCEDES TO SPREAD HIS BRAIN ALL OVER THE ADJACENT WALL, I DONT THINK HE WILL CARE WHICH BULLET HITS FIRST.

I KNOW I WOULDNT.
 
Are you assuming:
The wind will carry the bullet, much like water would a raft? Thus adding the velocity component of the wind to the velocity of the bullet?
The bullet is 672 fpm at the time it reaches the end of the muzzle and never slows down until it cracks the wig?

The first thing we need to do is put all of our numbers in common units and assign variables.

Gun1 = Right Hand
Vg1 = velocity of gun one
Xg1 = distance of gun one from wig

Gun2 = Left Hand
Vg1 = velocity of gun one
Xg1 = distance of gun one from wig

Let's use the metric system

Vg1 = velocity of bullet from gun one plus the eastern component of the wind
= (204*60 + 0.707*45000)m/hour
= 12240 + 31820
= 44060

Vg2 = velocity of bullet from gun two minus the eastern component of the wind
= (204*60 + 0.707*45000)m/hour
= -12240 + 31820
= 19580

Xg1 = .15875 m
Xg1 = 3.8 cm = 0.038 m
now we make use of kinematics

X = Xo + .5(V + Vo)t

We'll take Xo to be zero and solve the equation for t

t = 2X/(V + Vo);
since for our case V = Vo
t = X/V

So now solving for the time:

t1 = Xg1/Vg1
= .15875/44060 = 3.6 microseconds

t2 = Xg2/Vg2
= 0.038/19580 = 1.94 microseconds

So the bullet from Gun two reaches first which is the left hand gun

Damn, I probalby miscalcalcuated but I rushed becuasue it's time for me to go home!
 
Well, I have a problem with the 'both guns fire bullets at the same velocity'. That is bullshit. The longer the muzzle, the faster the bullet. That is why rifles are able to cause more damage. A bullet from a .22 pistol better be inches from the target, otherwise it will only piss the dude off. A .22 from a rifle however will do some damage.

But for the math problem, if both guns fire bullets at the same velocity and friction is not a variable, then the gun with the shorter muzzle will hit first, obviously.
 
Well, I have a problem with the 'both guns fire bullets at the same velocity'. That is bullshit. The longer the muzzle, the faster the bullet
Click to expand...

First mistake of novice students. Reread the problem.

gravity
Click to expand...

Gravity would have the same effect on both bullets, thus, it can be ignored.
 
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Troll said:


First mistake of novice students. Reread the problem.

Gravity would have the same effect on both bullets, thus, it can be ignored.
Click to expand...


I read the problem, did you read my answer? Reality is that the longer the muzzle, the faster the bullet. This problem was simplified for math students to derive an answer. I never said that gravity affects one over the other. I said the bullet from the gun with the longer muzzle would, in reality, travel faster.
 
BigPhysicsBastard said:
The question clearly stated that both bullets are traveling at the same speed, and everyone knows that the shorter the distance, the shorter the time taken to cover that distance given all things equal
Click to expand...


Yes, BPB, that is what I was trying to say... and I'll say it again:

"The one from the left hand gun

1.5" < 6.25"

assuming that the both bullets travel at the same rate of speed..... all other info is just B.S."
 
I read the problem, did you read my answer?
Click to expand...
Dude, it doesn't say anything about the length of the muzzle on the freaking guns.
Damn it, that's what I mean. READ IT. It describes the distance the muzzle is away from his head!!!
The left-hand gun muzzle is 1.5" (3.8 cm) from his left temple. The right-hand gun muzzle is 6.25" from his right temple.
Click to expand...
READ IT. IT DOESN'T SAY THEY ARE TOUCHING HIS HEAD, WHICH WOULD IMPLY ONE WAS LONGER THAN THE OTHER!
 
TxLonghorn said:
Well, I have a problem with the 'both guns fire bullets at the same velocity'. That is bullshit. The longer the muzzle, the faster the bullet. That is why rifles are able to cause more damage. A bullet from a .22 pistol better be inches from the target, otherwise it will only piss the dude off. A .22 from a rifle however will do some damage.
Click to expand...

I'd rather be shot with a .45 then with a .22. The .45 will just go through you, but the .22 is going to bounce around inside you.
 
WRONG Only if they are the different heights from the ground:eek:
gravity of the earth, if the guy is on earth, pulls downward. Toward the center of mass of the earth. We are indeed only concerned with the horizontal component of velocity of the bullets.
The only thing gravity will do in this case is make one hit him in the ear hole and the other in the earlobe.:D
 
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it does make a difference because the more it travels the more velocity it loses and the more gravity pulls it to the ground. i saw a show with snipers taking shots from a quater mile away, and they said that must take gravity into account. there would be some sort of difference due to gravity but it will be insignificant most likely in this case unless you want to get in billionth's of a second
 
Last edited:
good :cool:

now say "i'm sorry troll, you are a better reader than me. i being a texaslonghorn didn't learn how to slow down and evaluate a problem before jumping toward a conclusion. i wasted money on an expensive education when all i had to do was find you, and let you show me the way:alien: "

;) :D ;) :D ;) :D
 
Troll said:
good :cool:

now say "i'm sorry troll, you are a better reader than me. i being a texaslonghorn didn't learn how to slow down and evaluate a problem before jumping toward a conclusion. i wasted money on an expensive education when all i had to do was find you, and let you show me the way:alien: "

;) :D ;) :D ;) :D
Click to expand...


i am not sorry for anything, and i don't want a part of your education either. you can keep it to yourself. if you got an education to educate me, then you should ask for a refund. i said there would be a difference, and i am happy that you agreed.
 
I hate to break this to you troll, but I got the correct answer. I did misread the part about the muzzle length/distance from head.

Lol, you must not be used to reading to be so happy about that part, and also so distraught about it at the same time.

So, to answer yellowd's question, the answer is the gun that is closer to the head. The punchline is troll. :eek:
 
fist,
bro. i'm just trying to help you understand the world you live in. the bullet will not slow down due to gravity. i'm sorry, that's just the way it is.
V = Vo + .5*a*t^2
V is velocity.
Vo is the initial velocity.
a is acceleration
t is time
there are three components of velocity in a cartesian coordiante system. If we lable our axis X, Y and Z, there will be Vx, Vy, and Vz components of the velocity.
for this problem we let the positive X axis point toward the EAST, the Z axis points SOUTH, and the Y axis points straight up.
Now Gravity is 32ft/s/s in the negative Y direction.

This means ax=0, ay=-32/ft/s/s, and az=0.
so,
Vx = Vox + .5*ax*t^2
Vy = Voy + .5*ay*t^2
Vz = Voz + .5*az*t^2

plugging in our understanding of gravity we find that low and behold.

Vx = Vox
Vy = Voy +.5*-32*t^2
Vz = Voz

The components in the x and z directions are unaffected by gravity.

your thinking to pratical, of course the bullet will hit the ground eventually, due to the decrease in the Vy, but this problem doesn't address that portion of the motion.

*note: if you look at the equation for Vy you can see that time will always grow, so eventually the second part of that sum will become dominate, and will change the direction of motion.
 
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Ok. What if the length of the left side of his mullet is longer than the right side of his mullet? Will that throw it off?
 
gravity never really "takes over". no one wins, the fundamental law of Conservation of Energy says that all of the energy in a system remains in that system, it is simply transformed.

If we shot the bullet straight up in the air,(remember we are ignoring the effects of friction, which eliminates airdrag) when it comes back down, at the height at which it was shot, it will have EXACTLY the same velocity as when it was fired.

There are two forms of energy in a frictionless problem such as this. Potential and Kinetic. If we considered friction, we would have to consider the energy lost in the form of Heat. Kinetic energy is transformed in to Potential as the bullet rises and gravity causes the bullet to accelerate 9.8m/s/s in the opposite direction. Once all of the energy becomes Potential, the bullet will stop and begin it's decent toward the source of the accelration(earth's gravity).

When a sniper shoots "high" he is accounting for the effects of the vertical gravity. but the horizontal velocity never changes until the bullet actually strikes something, and at that point the energy is transformed to heat, and fucking up it's victim.

but when homey blows his brains out, we don't have to consider it because we are looking at the horizontal component of the bullets motion.We could actually ignore the wind too because the magnitude of it's velocity is significantly less that that off the bullet.
 
ok. i'm done.
fist,
the education stuff was for longhorn. just trying to create a pun from his name. i'm not trying to play smarter than thou and shit. i don't want to piss anyone off. i'm just having fun with my knowledge of physics, as a matter of fact when THeMaCHinE finally comes back, he'll probalby prove us ALL wrong.

both of you guys are right and i was just picking at insignificant details. i didn't mean to imply that i felt i was better than anyone else here, like i said, i was just dorking around.
sorry.
 
Travelling at exactly the same speed, the bullet fired from the gun closest to the head will hit first, which would be the left. Barrel length is irrelevant as the bulled is not fired until it leave the barrel, not when the trigger is pulled. It therefore leaves the gun at the same speed and the closest will hit first. Gravity is irrelevant over such a short space and with two bullets of equal mass. The only way gravity could effect the speed with no resistance as it stated is if the bullets were not fired at the same angle ie. one upwards and one downwards. However it states both guns are pointed at each side of his head, meaning they are both parallel to the floor.
 
I'm not going to work out the math problem here but note that both guns fire bullets that travel incredibly slow.
672 fpm = 11.2 feet per second or about 3.42 metres per second or 12.3 kilometres per hour...

You could throw a bullet faster than that...
So I guess the question is irrelevant because any bullet travelling so slowly would hardly leave a bruise on him anyways... (unless of course we're talking about depleted plutonium tipped super high explosive blow your friggin head off kinda bullets :D )

hardainer (being a real smartass right from the get go)
 
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