Ballistics !, Bullet question.?

Question

If I had a firearm and fired it absolutely horizontally from a height of twenty metres, how far would it travel before it :

a) decreased in speed,
b) decreased in height,
c) was victim to gravity and earths curvature,
AND
d) would it still carry enough force to injure somebody upon falling to earth ?

I know the calibre of firearm alters the answers, but if someone could post a general answer or some alternatives. Thank you.

Answer 1

a) It will decrease in speed as soon as the accelerative force of the explosive gas ceases to be greater than the drag of the gun barrel or the air or other media it passes through. This could mean that it begins to slow down in the barrel, or soon after leaving it. Certainly, after a few feet from the gun barrel one would expect the gases from the gun to be dissipated and the bullet to begin to be dragged by the air. However, the reduction in speed of a high velocity bullet could be negligible for seconds after firing.

b) As soon as the bullet leaves the barrel it will begin to drop towards the earth. The formula for this is:
distance =0.5*acceleration*(time-squared)

Acceleration at sea level is 9.81 metres per second per second. Thus it will take almost exactly 2 seconds for the bullet to drop the 20 metres from your firing platform. (square root of [20/(0.5*9.81)] )

c) See b above. The bullet is acted upon by gravity at all times. It is only prevented from being accelerated downwards by a force equal or greater to its weight pushing it up. In your example the gun barrel supports it until the bullet leaves the barrel, then begins to fall. Even if you pointed the gun straight up, the bullet will be slowed by gravity as soon as it leaves the barrel and gases behind, gradually slowing to a stop and falling back to land on your head.

If the bullet had enough horizontal velocity to make it to the horizon within two seconds, it would have acheived the required "escape velocity" to leave or orbit the earth, always falling but never landing. As the horizon at sea level is eight miles away this would need a speed of 28,800 miles per hour or 46,340 kilometers per hour.

The speed of sound is c.1,220 km/hour, so this would be 38 times the speed of sound!
The muzzle velocity of most military calibres is c.700metres/second or 2, 520 km/h or twice the speed of sound.

Thus in the two seconds it takes the bullet to fall to earth it would have carried some 1,400 metres.

d) Very much so!

Answer 2

A- it depends on the ballistic coefficient of the bullet. a more streamlined bullet will lose speed slower
B- it will fall at the same rate as one dropped with no forward speed
C- see A & B the curvature will not enter in to the equation since the bullet cannot travel far enough for it to become a factor
D- yes depending on the flight path and the range a bullet can easily still have enough energy to kill at the point where it strikes the ground.

Answer 3

a: It starts decreasing in speed the moment it leaves the gun's muzzle due to friction of the air.

b: It starts decreasing in height immediately.

c: Same as the answer to b. It is victim to gravity the moment it leaves the gun's muzzle.

d: Yes, it will still carry enough force to injure someone.

Answer 4

I'll answer with the assumption you are firing in a vacuum, since air resistance can not be figured out with your inputs...

A) It would not decrease in speed.

B & C) It would begin decreasing immediately. It would not go far enough to be affected by Earth's curvture, since it would hit the ground in just about 2 seconds.

D) It's horizontal speed would not decrease until it hit something, so yes, it would be able to injure someone.

Answer 5

If it was fired on a hypothetical airless planet (I know you said Earth) and the charge could be increased and increased, the bullet would only be affected by gravity and on each increase in charge it would go further and further until it went full circle around the planet and landed at the back of the firearm. if the charge could then be increased still further and immediately the firearm was removed from it's original position the bullet would be at orbital velocity. only the said hypothetical planet must have no hills higher than nineteen meters. A still further increase in charge would result in escape velocity.

Answer 6

a) Drag force in air is proportional to velocity. Speed starts to decrease immediately after leaves the gun.
b) decrease in height is proportional to square of time - small in the beginning, increasing with time.
c) this (victim of gravity and earth's curvature) is also immediate.
d) this really will depend on shape and smoothness of the bullet; one with a parachute, for example, will have all its forces spent.

Answer 7

It starts decreasing is speed due to air friction and height due to gravity immediately after it leaves the barrel. You need to know the density of the air, speed of the bullet and ballistic coefficient of the bullet to calculate the friction of air. The easy part is gravity. The bullet simply accelerates down at 9.8 meters per second per second all the time.

Answer 8

from the moment it leaves the barrel it starts slowing and dropping.Dependant on calibre and amount of charge, this will dictate the distance it will travel I know the following does not answer your question but it is interesting anyway. If you drop a bullet by hand at the end of a gun barrel it will land at exactly the sake time as the one fired by the gun.

Answer 9

a) Shortly after leaving the barely and the gasses pushing it dissipate;
b) as soon as it leaves the support of the barrel;
c) Ditto b
d) totally depends on the weapon fired - it would follow a ballistic trajectory and impact with a force dependant on the type and calibre of the firearm - a musket maybe 100 yds, a .50 call Barrat sniper rifle a couple miles!

Answer 10

a.infinite distance ,b,coz its speed will only go on increasing ,and air friction depends on the surface area of the object so that obviously wont effect that much.
b.zero ,as it will start decreasing in hieght once it leaves the fire arm
c.again zero b,coz it will be in the influence of the gravity once it leaves the fire arm.if its velocity is good it may even go into orbit
d.this question can be answered only if we know the caliber of the fire arm .if it is less it may break its projectile and start free fall