2006-07-26 09:14:22 · 16 answers · asked by jusme 1 in Science & Mathematics ➔ Mathematics
Depends on the initial muzzle velocity of the gun, which depends on the design of the gun and what kind of ammunition it is firing.
Muzzle velocities can be as low as 330 m/s (1080 ft/s) for low powered round like a .22, or as high as 1800 m/s (5900 ft/s) for supersonic tank guns.
2006-07-26 09:17:51 · answer #1 · answered by Anonymous · 0⤊ 1⤋
The formulae are
S = ut + 1/2 g t^2
v = u + gt
where S is the distance it would go up
u is the muzzle velocity
t is the time it stays up in the air
v is the final velocity (which is zero at the apex)
g = 9.8 m/s^2
As you see everything depends on the muzzle velocity which will change from gun to gun.
The weight of the bullet doesn't come into it (once the muzze velocity is known)
Basically, the faster a bullet comes out of the gun, the higher it will go.
Bullets typically have a muzzle velocity around 1 km/s (=1000m/s)
So using v = u+gt
0 = 1000 -10t (using g = -10)
t = 100 seconds
S = ut + 1/2gt^2 = 1000x100 - 5 * 10,000
= 50000
which comes to around 50 km.
But in practice it will be much lower than this because of air friction etc.
2006-07-26 17:31:49 · answer #2 · answered by blind_chameleon 5 · 0⤊ 0⤋
As above, it would require knowing the velocity at which the bullet left the gun. Making certain assumptions (such as constant gravitational pull, no wind etc), one could use the simple formuae;
v = u + at
where (v = final velocity (0)), (u = initial velocity), (a = acceleration due to gravity (around 9.8 m/s^2) and (t = time in flight)
Once one had the total time in flight, you could plug it into the following equation;
s = ut + 1/2at^2
where (s = vertical displacement), and the other variables are defined above to get your height.
Alternatively, one could derive the same result from;
v^2 = u^2 + 2as
Q.E.D
I only realised this once I'd written all the above out, however!
2006-07-26 09:30:24 · answer #3 · answered by sly` 3 · 0⤊ 0⤋
Use the formula:
v^2 = u^2 + 2as
You know v = 0 because at its highest point it will have no velocity.
u = the inital speed of the bullet from the gun, varies a lot depending on what gun, express in metres/sec.
a = -9.81 metres/sec/sec the deceleration due to gravity.
Solve the equation to s, which is the displacement (or distance travelled) in metres.
This of course doesn't take drag into account, that's beyond my knowledge.
2006-07-26 09:29:04 · answer #4 · answered by anonymous_dave 4 · 0⤊ 0⤋
It depends on the spped that the bullet started at.
From starting speed, the bullet would de-celerate at a rate 9.8m/s/s due to gravity, so it would slow down to a speed od 0, where it would stop.
if acc=(final speed - initial speed)/ divide by time, then time = acc/ final-initial. You then draw a graph to get the distance as the are under the graph.
Please re-ask the question telling us what speed the bullet was travelling at as it left the gun, or the acceleration rate of the gun.
2006-07-27 13:12:55 · answer #5 · answered by Anonymous · 0⤊ 0⤋
It definitely depends on the design, such as the muzzle, the bullet size, the type of gun, and the initial velocity. This question needs more specifying in order to give you a good answer. ;)
2006-07-26 09:23:25 · answer #6 · answered by Richard 3 · 0⤊ 0⤋
The german paris gun in world war 2 fired a round that went up to 40 kilometers in the air before it hit paris.
2006-07-28 02:17:55 · answer #7 · answered by scifi_fan69 2 · 0⤊ 0⤋
Depends on type of gun, type of bullet and the muzzle velocity of the bullet
2006-07-27 00:33:20 · answer #8 · answered by Paul B 5 · 0⤊ 0⤋
Dependent on mass of bullet, speed of emergence from the gun itself.
2006-07-26 09:21:42 · answer #9 · answered by Fi 2 · 0⤊ 0⤋
It also depends which planet you are on as (de)acceleration due to gravity comes in to play. Friction due to air resistance of the bullet would also need to be estimated. I can't remember the linear motion formulae from my physics lessons cos it's been 12 years.
2006-07-26 09:27:10 · answer #10 · answered by Quasimojo 3 · 0⤊ 0⤋