Flat Trajectory?

Question

A bullet shot from a very high velocity rifle may travel one hundred feet or more without dropping at all.

a) True
b) False

The answer is: b. There is a popular idea around, even taught in some police academies, that a sufficiently high-powered bullet will go some distance without dropping at all. But this is a real misconception. The speed of a thing cannot "turn off" gravity, not even for a moment. Even light from a laser begins to drop the moment it begins its flight. Moreover, the rate of drop is always the same, regardless of speed. If the bullet is fired horizontally in a so-called "flat trajectory," it falls 16 feet during the first second of flight. However, during that second a high speed bullet will go farther than a low speed bullet so the flight path or trajectory of the high speed bullet looks less curved than the trajectory of the low speed bullet, but both always curve. A trajectory can never be completely flat (unless it is straight up or straight down).

Answer 1

Definitely false! Regardless of muzzle velocity a bullet begins to fall the instant it leaves the barrel of the gun. If the barrel is horizontal a bullet dropped from the height of the muzzle will hit the ground at the same time as the fired bullet, higher muzzle velocity will cause the bullet to travel farther, that is the only difference.

Answer 2

The simple answer, is baring some outside force acting on the bullet, or it achieving orbital velocity ~8.0 km/sec., it will drop. Simple d=1/2a(t)^2 and t is never zero. Even simple aerodynamics won't help because, in order to get an angle of attack, the bullet will have to drop.

But, love to be a contrarian, so I'll assume "A" for the moment.

a. Just assume that the rifle is pointing straight up. The bullets not dropping, its going up.

b. As Kirchwey points out, Magnus force and a crosswind.

c. The bullet exits the rifle at orbital velocity. (Maybe on the moon where orbital velocity is less).

d. The bullet exits the rifle with an angle of attack above the direction the rifle is pointed and causes immediate lift.

e. The bullet is a ball, and it has backspin.

f. The bullet is fired in a medium that has the same density as the bullet (it won't sink).

g. The bullet is fired into a medium of variable density (or variable velocity, i.e., wind shear) such that, even though the bullet is going straight, there is more lift on the top side.

h. There is a small up component to the wind.

i. Frame of reference. Both the shooter and the bullet are in the same frame of reference and experience the same forces, i.e., both are in free fall.

Since all of the above don't really work well or are a bit unusual, we are stuck with "b."

Answer 3

The fastest bullets typically leave the barrel of a rifle at about three times the speed of sound - lets call it 1000 m/s to keep the numbers simple.

100 ft is equal to 30.4 meters, so that neglecting air friction, the bullet would cover the distance in 30.4 / 1000 seconds, or 30.4 milliseconds.

We know that gravity is always pulling the bullet downwards - but it starts to fall when it leaves the barrel. So we can apply s = ut + 1/2 g * t ^2. For now, we can assume the initial velocity is zero so that s = 1/2 * 9.81 * 0.0304^2 (meters). I get 0.00453 meters, that's 4.53 millimeters or 11/64 inches.

So the bullet drops a fair bit from the straight line flight.

Some things to think about - firstly the greatest drag happens as the bullet leaves the barrel so that its velocity is not constant - this means it is in flight for longer and the drop is slightly larger than the calculation suggests. Secondly, bullets are never fired horizontally - they are usually fired slightly upwards so that their trajectory is effectively longer.

Last thought - an anti-tank gun can have a muzzle velocity up to mach 6 - or around 2000 m/s. The projectile is extremely heavy so that it doesn't deccelerate much and it has very little cross sectional area. These weapons really do have a flat trajectory.

Answer 4

For all practical purposes the trajectory is flat. But we all know that gravity acts on the bullet. One hundred feet is a very short distance for a high velocity bullet to travel. The hole it makes would be larger than the drop. So answer A if this is for ballistics; B if its for physics.

Answer 5

Grasping at straws. Assuming 0 initial vertical velocity, a spinning bullet, and presence of aerodynamic effects; also assuming, considering the asker, that there's a catch somewhere.
How about the Magnus effect? This provides lift to a rotating cylinder with a sideways component of motion through the air. Could a bullet tumble enough to benefit from this effect? Not likely in the first 100 feet. But a sidewind could do it. So I'll vote for A.
EDIT: I found a reference to the Magnus effect on a bullet. Apparently it is common for bullets to have a yaw angle in flight, which would produce the equivalent of a sidewind. The example the ref. gives results in a downward Magnus force, but if a sidewind were strong enough and in the opposite direction, force should be upwards.

Answer 6

B. The bullet starts to fall as soon as it leaves the muzzle.