What happens to a bullet shot straight up in the air? It does not go into space. Of course, it falls back to earth. The details, however, are interesting:
A bullet leaves the barrel of a gun at 2,000 – 3,000 feet per second. The bullet leaves the barrel nose first and spinning around its axis, which provides stability, due to the spiral grooves in the gun barrel.
When a bullet is fired 90 degrees to the horizontal the bullet is fighting the force of gravity and will gradually slow until it stops going up (at about 10,000 feet – or 2 miles) and then will fall back to the earth. If there were no air resistance, the bullet would return to earth going exactly the same speed as it left the gun barrel, i.e. around 2,500 feet per second. This is because gravity is a constant and gravity will cause the bullet to accelerate on the way down at the same rate as it slowed the bullet down on the way up.
However, on earth there is air resistance. As the bullet falls air resistance will cause the bullet to stop accelerating and hit terminal velocity (terminal velocity is the speed at which air resistance balances the accelerating force of gravity). Experiments have determined that falling bullets reach terminal velocity at 200-300 feet per second depending on type. Note that falling bullets (shot vertically) usually do not come down nose first – which would be the most aerodynamic – but instead tumble, which really slows the bullet.
When you fire a bullet into the air, it typically takes between 20 and 90 seconds for it to come down, depending on the angle it was fired at, its muzzle velocity and its caliber. So, if you are a bystander, you have some time to take cover.
A few questions this may raise and some answers:
- Can being hit with a bullet traveling 200-300 feet to second kill or injure you? Maybe. A bullet traveling at that speed might penetrate the skin depending on where it hits you. There are cases of people dying after being struck by falling bullets and other cases where there was only slight injury. But, most bullets shot up in the air are not shot exactly 90 degrees vertical and adding horizontal component to the firing of the bullet will increase the terminal velocity speed as a bullet shot at an acute angle maintains a ballistic trajectory and is not likely to engage in a tumbling motion. So, actual cases of injury or death might only be reflective of bullets fired at an angle other than 90 degrees to horizontal.
- What are the chances of the bullet that you shoot straight up coming back down and hitting you? Nearly zero. First, it’s very hard to shoot completely vertically. Second, wind will likely push the bullet far from the location where it was shot. Third, the spin of the earth will have a slight effect on where it will land compared to where it was shot.
Look into the coriolis effect
What happens to the rotatory motion of the bullet (due to the Groves) when coming down from a vertically upward shot?
Indeed Interesting FOD😊
I’m questioning the statement “If there were no air resistance, the bullet would return to earth going exactly the same speed as it left the gun barrel, i.e. around 2,500 feet per second.”
After the bullet from the gun barrel stops (at about 10,000 feet), wouldn’t it fall back to the earth to the same speed as it were just let go down from the same height (about 10,000 feet), which has nothing to do with the “speed it left the gun barrel”? Could someone clarify?
Alex- you are correct – it would be the speed at which you’d drop it. Which would be the same speed it left the barrel because the effects of gravity would be symmetrical.
Michael – my point was that the dropped bullet doesn’t have any association with the speed with which a bullet leaves a barrel, i.e. it would be determined – as you confirmed – by the effects of gravity, so that the statement “the bullet would return to earth going exactly the same speed as it left the gun barrel” is incorrect – isn’t it?
Think about it this way: a bullet leaves the barrel of a gun at 2500 ft./s. When it is shot straight up if we had no air resistance the bullet would slow based on the force of gravity at 32 ft./s squared. After thousands of feet gravity will have slowed the bullet’s upward movement to the point of apogee. Now, the bullet will fall back down, again without air resistance, accelerating at 32 ft./s squared. Thus, it would return to its starting point at the same speed that it left the barrel of the gun
Oh, yes, thank you!
This is totally incorrect! Going upward a bullet has 3 forces working on it. An explosive force (from firing the gun) that accelerates it to 2500 ft/sec, gravity (opposing force) and air resistance (also opposing force). Air resistance is constant whether going up or down so it can be cancelled out in comparison.
At the bullets apex where it slows to zero ft/sec due to opposing forces, it will start to reverse direction and head back to earth from again, zero ft/sec.
On the way down the bullet is acted on by only 2 forces – not 3; gravity (this time as an accelerating force) and air resistance (which is constant going up or down so it can be cancelled) BUT no explosive force to accelerate it to 2500 ft/sec so it will be traveling no where near 2500 ft/sec as you describe. It will only have gravity to act on it to bring it to terminal velocity so from 200-300 ft/sec depending on the size/weight of the bullet.
Do the math. Ignore air resistance for now. If an object is traveling at 2500 feet per second and is being slowed by gravity of 32.17 feet per second squared, it will take 78 seconds to slow the bullet to a stop. Without air resistance, the bullet will be 97,862 feet in the air (that’s high!). Then just reverse it. A bullet with no intial velocity that falls 97,862 feet will have an ending velocity of 2500 feet per second. Here’s a website with as useful calculator of these things:https://www.omnicalculator.com/physics/free-fall
Of course there is air resistance that changes what happens going up and down since the bullet is aerodynamic (but still subject to air resistance) on the way up but tumbles on the way down (so has a ton more air resistance).
On the way up, only two forces are acting on the bullet: gravity and air resistance. The explosive force is a given (without it the bullet would just drop) or think of the problem as “how far into the air will a bullet go that is already traveling 2500 feet per second). Then on the way down, same two forces: air resistance and gravity. Of course, as explained, the bullet has a greater air resistance coefficient on the way down due to it tumbling.
And if you still don’t believe me — this IFOD has been read by multiple PhD physicists who agree.
Basically, the speed at which the bullet leaves the barrel determines how high it will go, and therefore be dropped from.
It would not be the same unless the bullet has another explosive force at its apex to speed it up from zero ft/second. It would indeed only accelerate as much as if you dropped it from the highest point.
You are correct, the bullet would not hit the earth at the same speed that it left the gun barrel, it would return at “Terminal Velocity”, the fastest an object can fall in the air. US Army General Julian Hatcher did a detailed study of this in a publication called “Hatcher’s Notebook”, published 1966, which is available on the Web for free. It’s a very interesting read.
The earth spins at around 25,000 mph….. So eventually the bullet would lose the momentum it originally had like throwing a ball out of the moon roof of your car, at first it’s in sight but we in the car have constant propulsion, (gas pedal) the ball loses its drive to keep up with the car…
It’s going to fall fast and hit the ground, but not right where it was thrown up… (Simply put)
And without atmosphere and air and wind, the bullet would fall at exactly the same speed. It’s also going to start to tumble on the way down, slowing it even more. The barrel of the gun is designed to spiral the bullet, giving it an aerodynamic advantage over the one falling….
But that’s not the EVER going to be the case on earth. So it’s only a technicality, not a reality..
CORRECTION!!!!
I’m sorry I was speaking into the phone and not typing, so when I said the Earth spins at about 1,000 mph,
SOMEHOW the phone heard me say 25,000 mph.
According to my knowledge
It will continue accelerating until it reaches the speed of light (which is impossible) , thats why it impossible for the universe to exist without air resistance
As you know, I minored in physics. The explanation from Joe confirms that I made the right decision not to try to major in physics.
All the best for Happy Holidays my friend. Keep up the blog!
First, thanks as always. Such a great detour in my day. Second, quick question about the third point in your second question above. Why would the rotation of the earth have any effect on the bullet, given that (in an Einsteinian way) the bullet and the earth are at rest relative to each other, and (except for the fired bullet’s upward and then downward motion), they stay that way? In other words, the bullet will have the same “forward” or “orbital” momentum as the earth does during its entire trip and should, all other things being equal (i.e., no air resistance, perfect vertical angle), come down at the same earthly coordinates unaffected by the earth’s spin. I think. But it’s Friday and it’s been a long week, so who knows?
Great question Chris -I could not answer it so I asked Joe Incandela, as physicist at UCSB. Here’s his answer:
I will try to explain without equations. I hope it is clear!
The trick with this kind of physics problem is to view the earth from the vantage point of an external observer who is say, fixed in position relative to the distant stars, and so sees the earth as a spinning sphere and sees the point at which the bullet is fired as a fixed point in space, not a fixed point on the earth. So here goes…
It is true that the bullet has a component of its velocity toward the east – i.e. in the direction toward which the surface is rotating. Would this allow the bullet to always stay aligned above the same point on the surface of the earth from which it was fired? The answer is no because the part of the earth’s surface from where the bullet was shot is accelerating – it’s moving along a circular orbit about the spin axis of the planet. In other words it is “falling away” from the fixed point in space where the bullet was fired. Now you may say “but that’s ok because gravity is pulling on the bullet too and so its velocity is constantly shifting to stay along the eastward direction as seen from the point on the earth where it was fired”. This does not work for a couple of reasons. One is simply that as the bullet rises above the surface of the earth, the horizontal speed it would need to keep up with the surface must increase and there is no force along that direction, so the bullet is out of luck. In addition, as the bullet goes upward, the gravitational pull it feels gets slightly weaker, so the deflection of the bullet’s “eastward” motion is diminished and so it cannot stay along the eastward direction as seen from the place where the bullet was fired. As a result of all this, the bullet will land slightly to the west of where it was fired.
Earth is flat.. stop smoking that stuff man.
Many physics problems are solved with layered approximations. Here I provide 3 layers that should provide a better approximation than a simple cartesian frame of reference.
1)Vertical Frame: As the projectile goes up, the radius of the world at the height of the projectile increases, while the bullet has the rotation conserved from lower down, causing the bullet to curve relative to the rotating frame of reference below it ( the earth). The reverse happens on the way down, so when and if these forces overcome the stabilizing effect of the spinning bullet, coriollis forces/precession will differ on the tip and tail of the bullet, and it will try to tumble one way on the way up and another on the way down. Since the speeds up and down are different, the symmetry of the corriolis force is broken, and the sums will differ.
2) Pole/Equatorial frame: #1, simplest case, happens at equator, only in a vertical sense. However, if you are not on the equator, a similar mechanism (coriollis forces, the change in orbital height caused by moving towards/away from the equator and away/towards the poles ) also comes into play, causing a right handed force below the equator and a left hand twisting force above the equator. This can be seen in water draining from a bathtub, it generally swirls clockwise / ccw based on it’s hemisphere.
3)Solar Frame: #2 will be repeated with reference to the sun and the earths tilt relative to the sun. If the projectile is shot towards or away from the sun, a “solar coriollis” force will also affect the projectile. If shot forward or back relative to the earth orbit, solar coriollis forces will be negligible, but there may be effects related to shooting directly ahead then running into the bullet, vs shooting behind, and the bullet chasing the fleeing planet. Seems like this effect gets closer and closer to an orbit and less and less of a straight line, and the chasing bullet wants to go straight, but the force of gravity keeps “turning” to stay in orbit. Obviously that “turning” is the force of the sun’s gravity which would affect the planet and the bullet equally as the mass term cancels. #3 solar coriollis forces would be orders of magnitude less than the first 2.
Probably more detail to the model than needed, but was fun to think about.
You are totally right, and what a great way to start the year with some thoughtful analysis. Boy, life is more complicated sometimes than you think.
Love the snowflakes. Happy Holidays, John!!