Conquer Club

hi guys i was doing a physics problem and got an extreme answer and wanted to know what u thought.

Problem, a Gun has a muzzle velocity of 925 m/s (meters per second not miles per second), how far would it go if the gun was pointed at a 45 degree angle.

what i did

first we need the time it will fly (which is until it hits the ground) which is called T
Initial velocity in the upward direction = Vyi
Final velocity int he upward direction = Vyf
force of gravity = -9.80 m/(s^2)
it will hit the ground at the opposite speed it went up at so Vyf = - Vyi

Vyf = Vyi + -9.80 m/s^2 * T
-Vyi= Vyi + -9.80 m/s^2 * T
plug in
-925*sin(45) = 925*sin(45) + -9.80 m/s^2 * T
solve for T and get T = 133sec

now that we know the time we can just solve for the distance in the horiz direction by multiplying Velocity by Time
Time = T
Velocity in horiz direction = Vx
distance = D

D = V * T
plug in
D = [925*cos(45)] * 133 sec
D = 86991.81176 meters
D(with sig figs) = 8.70 * 10^4 meters
...................=54.1 miles
...................=87.0 kilometers.


there is no answer in the back of the book, but does that seem like a reasonable number to get?
-Adith

I get the same answer as you in a slightly different fashion, I believe it is correct. As far as I know its the muzzle speed of the bullet in the problem that's unrealistic.

just looked it up and that bullet speed is perfectly ok, so......

*edit* the math works on that answer, think it may not seem realistic since we're neglecting air resistance*

sfhbballnut wrote:just looked it up and that bullet speed is perfectly ok, so......

You are not taking friction/air resistance into account.

Your numbers would work if you were in a huge vacuum.

lol so the answer is realistic, but the 54 mi long vacuum cleaner isnt? got it ill put that as my answer

thx all for ur help

Well, the muzzle veolicity is almost 10% of escape velocity at sea level, so if you could ramp up the start speed by 10x - 12x it would never come back down.

incidentally, if you do ignore air resistance, the initial angle of the velocity should make no difference.

I think.

Try it with a number of different angles and see if I'm right.

jonesthecurl wrote:Well, the muzzle veolicity is almost 10% of escape velocity at sea level, so if you could ramp up the start speed by 10x - 12x it would never come back down.

incidentally, if you do ignore air resistance, the initial angle of the velocity should make no difference.

I think.

Try it with a number of different angles and see if I'm right.

haha well with more gunpowder maybe i could make the next sputnik a bullet

the angle does make a difference because it changes how much of the force is uses horizontally and vertically. the two extremes being 90 degrees where its all vertical, and 0 degrees where it is all horizontal.

Yeah. I think you're about 50 miles beyond what a normal bullet might travel.

This guy might know.

Damnit... I'm a physics major, and always ready to help someone with a problem like this, but currently I am intoxicated and about to go to bed, so I am unable to answer. If this issue is still open in about 8 hours when I wake up, I'll post whatever I can figure out.

No Gravity Gun pictured?

I'm ashamed.

Dude ... 925 METERS per second is high....here is a list of velocities for you to check out

http://www.recguns.com/Sources/VIIE8.html

So.....typical velocities for commonly owned guns would be 1000 fps (305 meters) for pistols and 2500 fps (762 meters) for rifles.

Also, wouldn't the weight / size of the bullet be a significant factor? The slow, heavy, fat 45 ACP typically weighs twice as much as the narrow, lighter, pointy 7MM....

Lastly, the largest, nastiest rifle (50 BMG) will not travel more than several (4) miles, but CAN achieve a muzzle velocity of what you used originally, so this can be a good control argument (but this round is not realistic to use, I do not believe)......

Finally, as it has been 25 years since I was in college (lol), I cannot decypher your calculation....primarily the final one where you use cos......is that how you are adjusting for an "average" speed of the bullet? Obviously, at landing (and I believe you addressed this), the bullet will be travelling relatively slowly....

50+ miles is completely impossible in any scenario.....so.....I look forward to your new info ;o)

A "bullet" from one of these would get there.

http://www.worldwar1.com/foto/fww2703.jpg

KoolBak wrote:Dude ... 925 METERS per second is high....here is a list of velocities for you to check out
...
Also, wouldn't the weight / size of the bullet be a significant factor? The slow, heavy, fat 45 ACP typically weighs twice as much as the narrow, lighter, pointy 7MM....
...
50+ miles is completely impossible in any scenario.....so.....I look forward to your new info ;o)

Heh... well, physics problems often include unrealistic things, just for hypotheticals. I can't tell you how many problems I solved back in phys151 that had a block pulled along a "frictionless surface" by a "massless string" attached to a "frictionless, massless pulley." So in a.sub's problem, the question is: "this gun shoots a bullet at 925 m/s. It just does; maybe it's a futuristic super-gun or something. But in any case, the bullet starts at 925m/s at 45° and at ground level. How far does it go?"

a.sub wrote:hi guys i was doing a physics problem and got an extreme answer and wanted to know what u thought.

Problem, a Gun has a muzzle velocity of 925 m/s (meters per second not miles per second), how far would it go if the gun was pointed at a 45 degree angle.

what i did

first we need the time it will fly (which is until it hits the ground) which is called T
Initial velocity in the upward direction = Vyi
Final velocity int he upward direction = Vyf
force of gravity = -9.80 m/(s^2) acceleration by earth's gravity should either be g = 9.8, or if you need another decimal place, 9.81. Not that it really matters that much for this problem.
it will hit the ground at the opposite speed it went up at so Vyf = - Vyi

Vyf = Vyi + -9.80 m/s^2 * T
-Vyi= Vyi + -9.80 m/s^2 * T
plug in
-925*sin(45) = 925*sin(45) + -9.80 m/s^2 * T
solve for T and get T = 133sec
This all seems right so far.
now that we know the time we can just solve for the distance in the horiz direction by multiplying Velocity by Time
Time = T
Velocity in horiz direction = Vx
distance = D

D = V * T
plug in
D = [925*cos(45)] * 133 sec
D = 86991.81176 meters
D(with sig figs) = 8.70 * 10^4 meters
...................=87.0 kilometers.


there is no answer in the back of the book, but does that seem like a reasonable number to get?
All your work there is correct, so that should be the correct answer to the physics problem. There may not be a gun that actually fits the conditions in the problem (also, how often does someone shoot a gun at 45°?), but homework problems for practice are all hypothetical anyway.
-Adith

HapSmo19 wrote:A "bullet" from one of these would get there.

http://www.worldwar1.com/foto/fww2703.jpg

Now SOMEBODY must have been wicked insecure to buy that.

spurgistan wrote:

HapSmo19 wrote:A "bullet" from one of these would get there.

http://www.worldwar1.com/foto/fww2703.jpg

Now SOMEBODY must have been wicked insecure to buy that.

just because im paranoid doesnt mean nobody is out to get me!

TheProwler wrote:

sfhbballnut wrote:just looked it up and that bullet speed is perfectly ok, so......

You are not taking friction/air resistance into account.

Your numbers would work if you were in a huge vacuum.

Word. That plus the fact that the speed of the bullet is way too high compared to a normal gun and you're bound to end up with weird numbers.

He's doing a practice problem that demonstrates how gravity affects a projectile. You need to learn about this stuff, then learn how to factor in air resistance.

And that is the distance a bullet would go if shot from a gun that quickly (even if such a gun doesn't exist), if friction disappeared temporarily.

Did you take into account how far off the ground the gun is when it fires? It would only make a small difference, but a difference nonetheless.

well thx for the help!
and here are some quick things i found
1) M16 muzzle V is 975 m/s (wikipedia.org) so its not all that unreasonable, although giving me one of those and letting me shoot it is.
2) the gun is on the ground
3) no air friction sorry, right now we are breathing out of scuba masks while shooting this in a plane with no air

Well I have 3 rifles that fire the .223 (M16 round) so next time you be up in Oregon, we'll head out to the desert and do some shootin ;o) But having sighted them in, I can assure you that if fired horizontally, approx 5 ft above the ground, it will hit the earth after several hundred yards. At a 45, it will go maybe 2-3 miles.

Now if you can find the low-gravity, vacuum environment you are looking for, it should go farther ;o)

FYI...here are some interesting bits of info re: the .223

http://www.thehighroad.org/archive/index.php/t-1811.html

http://www.thehighroad.org/archive/index.php/t-26142.html

http://www.answerbag.com/q_view/7153

OK....a little more (sorry...lol)

I am no rocket scientist, but I am trying to back into some calculations here and compare them with a special I have seen on tv re: the study of bullets shot straight UP.....those on average took less than 10 seconds to return to the earth and were travelling at terminal velocity, lets say around 120 MPH.

Assuming your bullet leaves the barrel at 3000 FPS and returns to earth at 150 FPS (Term Vel), lets say the average travel speed is 1575 FPS. If we KNOW the bullet will travel approx 2.5 mies (fact), then backing into the TIME, it comes up to a flight duration of 8 seconds, which sounds right. Interesting....now lets see you put it into the right equation.....lol ;o)