# How can a horizontally fired bullet reach the ground the same time a dropped bullet does?

I studied projectile motion and now I know that we can treat each component of motion independently. Since gravitational acceleration acts on both a horizontally launched bullet and a vertically dropped bullet in free fall, they both will reach the ground at the same time as their vertical initial velocity is zero. This is what I studied in high school. But I found it against a real observation that a horizontally fired bullet will travel for much longer time compared to a simply dropped bullet before hitting the ground. Could you please elaborate on how to connect the physics of the situation and real life observations?

• "But I found it against a real observation that a horizontally fired bullet will travel for much longer time compared to a simply dropped bullet before hitting the ground." You're going to have to elaborate on this, because it is inconsistent with most observations. To give you a well produced and vivid demonstration see the mythbusters episode where they explicitly tested this. May 10, 2018 at 0:11
• The simplistic view of the motion of the bullet does ignore turbulence and aerodynamics and several other factors, but they are (at our distance) mostly random factors whose effects cannot be readily summarized. May 10, 2018 at 0:37
• It is worth noting, also, that the intro class version ignores the effects of drag. However, for Rayleigh drag (proportional to $v^2$) you recover the fall time equality for dropped and horizontally shot projectiles. Rayleigh drag is not a bad approximation for heavy, subsonic bullets. May 10, 2018 at 1:00

Since I'm impatient I'll suggest one way you could be surprised: if

• You are comparing the carry time of a rifle bullet to a dropped bullet and
• The rifle sights have been zeroed in for non-trivial distances

then the barrel is not level when aimed at a target the same height at the firing point, but instead points slightly upward accounting for the observation handily. Indeed, it must be that way because if the bullet was truly fired horizontally then it can only hit targets lower than the barrel.

• You also need to shoot due east and due west to cancel the Eotvos effect (easy) and with right handed and left handed-rifled barrels to cancel the Magnus effect (hard)...or use a musket.
– JEB
May 10, 2018 at 2:39

Much of our intuition about physical movement is based on our life-long observations of things like darts, arrows, or balls. But a bullet has two properties that run counter to our intuition. (1) A bullet almost always has a C/P (center of pressure) forward of the C/G (center of gravity) which is the opposite of what we see with most visible projectiles. And (2) a bullet is spinning. Fast. Really fast, like 120,000 to over 300,000 RPM, so it has a lot of gyroscopic energy, which itself is very counter-intuitive.

So... Two phenomena at work here...

(A) When I drop a bullet, it almost instantly flips base-down and lands on its tail, so most of its fall is in that lower-drag configuration. A fired, spinning bullet will tend to spend most of its flight horizontal to the ground, and so the vertical component of its fall to the ground is on the higher-drag SIDE of the bullet, not the base. This drag won't become meaningful unless the height above ground is sufficient for measurable drag differences to occur closer to terminal velocity.

(B) Gyroscopic physics acting on a CG-aft-of-CP bullet induces some counter-intuitive shenanigans on our bullet's flight. Gravity is pushing down on the CG of the bullet. A large percentage of that force is pushing the bullet down to the ground. But wind resistance pushing up on the bullet as it accelerates downward acts on the CP and this causes a tiny amount of leverage acting on the CP-to-CG arm, continually trying to push the bullet into a nose-high, tail-low configuration. The spinning causes gyroscopic forces to translate this small fraction of gravitational forces from being downward toward the ground, to being right-ward (assuming right-twisting rifling, which is typical). Thus some of the gravitational force is diverted horizontally and is not acting to push the projectile downward. At 1,000 yards my bullet drop may be on the order of 240 inches and my spin drift may be around 8 inches... in which case about 1/30 of the gravitational force was diverted right-ward instead of downward. 1/30th of a one second time of flight is a measurable difference.

With both (A) and (B) you have to be dropping from tall heights and shooting at long ranges before the differences will be measurable to all but those in instrumented lab conditions. But the dropped bullet will indeed generally land a tiny time interval before the fired spinning bullet.

I can think of at least 3 other phenomena that would be part of a really thorough analysis, two of which others have mentioned. Eötvös could alter time aloft as the fired bullet would have potentially different centrifugal forces acting on it than the dropped bullet. At 1,000 yards, we may be altering the 240-inch drop by 4 or 5 inches, again, a measurable difference plus or minus. Magnus would affect the fired bullet in crosswind conditions, and could slightly increase or delay the fired bullet's landing. In a right-spinning bullet, Magnus slows the bullet's descent if the wind is from the left, accelerates it if the wind is from the right. Magnus would not affect the dropped bullet if it wasn't spinning at high RPMs. And finally, there's ground effects. As the fired bullet gets within inches of the ground, the displacement of air by travelling at 2500 feet per second is significant, and air displaced downward would create a tiny high pressure zone under the bullet once close enough to the ground, again holding the fired bullet aloft momentarily. This air displacement would be microscopically small in the falling bullet.

Hope this helps.

It happens in real life just as physics says.

It was tested on the TV show Mythbusters: https://www.youtube.com/watch?v=tF_zv3TCT1U