Will a bullet shot straight up travel as far as a bullet shot horizontally? For the horizontal shot assume the bullet is shot at whatever angle maximizes the distance traveled.
I think that the bullet will not travel as far vertically as it will horizontally. I don’t know why I think that.
Don’t let me down guys!
the vertical distance will be a lot less than the horizontal distance (assuming there isnt a ground to stop it prematurely). Force of gravity during vertical travel is far greater than the aerodynamic drag on the bullet when its traveling horizontally, thus the travel distances will be lower when fired vertically.
I am too stupid to answer this (ie actually figure things out) but remember that gravity and air resistance affect both shots.
I think the vertical shot will be shorter (only talking about the travel directly up, not when it runs out of steam and starts falling).
When the gun is fired horizontally, there is the drag force hindering its travel in the x direction while gravity pulls it toward the earth. In other words, the forces are perpendicular to each other, meaning that the drag force is really the limiting factor on the bullet’s distance (assuming the bullet can fall forever.)
When you fire a bullet straight in the air, the drag force and gravity now act in the same direction, resulting in a much greater resistance force to the bullets travel.
This might be oversimplified, but there’s no easy way to prove it or test it.
Since when does the law of gravity change due to orientation of the mass’s travel (at realistic distances from COE) or are you saying that the force of gravity is more of a factor in it’s “flight time” than the aerodynamic drag? Either way you shoot it though, it’s gotta cut thru the air regardless of if it is vertical or horizontal.
If you’re only accounting for flight time to the apex of travel, I’d say that the vertical shot would cover more distance, but if accounting for maximum flight time including the back half of the trajectory (falling), I’d say a 45degree shot is going to cover more ground. But simply considering it from a “flight time” point of view, the vertically fired bullet will spend far more time in the air than one fired perpendicular to the ground on a perfectly flat surface. Dammit, now I’m gonna have to go digging for my books again!
It’s been a while since high school physics (in which I got an A, if I recall ;))
Let’s take a muzzle velocity of 425 meters/sec (handgun), but it should be the same for any velocity:
Note: these calculations don’t take into account air resistance, earth curvature, wind, etc. etc.
Shot straight up, it’ll take t seconds to reach a velocity of 0 m/s (where it stops going up and starts going down)
t = v/g
v is initial velocity (425m/s) and g is gravity (9.812865328m/s/s)
t = (425m/s) / (9.812865328m/(s^2))
t = 43.31048942s
And another t seconds to come back down. (This time v would be final velocity)
Total time = 2t = 86.62097885s
Now, distance traveled in that time is average velocity (1/2* 425m/s = 212.5m/s) multiplied by time (86.62097885s) = 18,406.958 meters.
Shot at an angle to optimize distance (45 degrees) you have two components: horizontal velocity (used to calculate distance traveled) and vertical velocity (used to calculate time to drop). They will be the same.
These form a triangle where the hypotenuse is 425m/s. Remember the Pythagorean theorem: A^2 + B^2 = C^2. And also that in a 90-degree triangle (where the angle we care about is 45 degrees), the base and height are equal.
C = 425 m/s
C^2 = 180,625
A^2 = 1/2 * 180,625 (A = B, A^2 = B^2)
A^2 = 90,312.5
A = Sqrt (90,312.5)
A = 300.520382 m/s
B = A
Now we know how fast it is going up (to cacluate time in air) and how fast it is going forward (to calculate how far it goes)
Same calculation as before, different numbers:
Time in air = 2t
t = v/g
t = (300.520382m/s) / (9.812865328m/(s^2))
t = 30.62514077s
2t = 61.25028154s
Now, total (horizontal) distance traveled = t * average velocity (300.520382m/s)
Total Distance traveled = 18,406.958m
The exact same distance.
If you want to compare VERTICAL distance shot up and HORIZONTAL distance shot at 45 degrees, they’re the same.
If you want to compare VERTICAL distance shot up and VERTICAL distance shot at 45 degrees, the one shot upwards goes further.
If you want to compare TOTAL distance traveled, the one shot in an arc (45 degrees) travels further.
Now, even though the bullet is in the air for a shorter period of time (61 vs. 86 seconds) it is traveling (in the direction we care about) at a faster average speed (300 vs. 212 m/s). That is why the numbers come out the same.
Now, if you want an answer if you INCLUDE wind resistance, my guess is that the 45-degree shot will travel a shorter distance since it is in the air longer and therefore wind resistance will have a greater accumulated effect. Never learned that, though ;).
As far as a cool experiment, check this out from Mythbusters
I was referring to a horizontal shot with the barrel oriented at 0 deg compared to a vertical shot with the barrel at 90 deg. I missed the part where the OP stated that is would be at an angle to maximize the distance.
When the barrel is at 0 deg, the force of gravity will be acting directly downward, or 270 deg and the drag will be acting in the opposing direction of the bullet’s travel, or 180 deg. When the barrel is at 90 deg, both the gravity and drag force will be acting downward at 270 degrees. Since the acceleration of gravity is constant and the drag force is related to the velocity of the bullet, they should not change with the barrel’s orientation. Basically, the forces will now “stack” on each other, slowing the bullet faster.
we need some clever person to add air resistance to this equation. looks like you’re on to something so far.
First thoughts on air resistance:
It decelerates the bullet so the longer the bullet is in the air, the more it should slow it down. The problem is complicated, however, by trying to figure out drag coefficients and all that jazz.
But on second thought:
You’ve got terminal velocity issues. I calculated the vertical component of the 45-degree shot at 300 m/s. I’m fairly certain that the terminal velocity (the point at which the force of drag = the force of gravity and the bullet no longer accelerates but reaches its maximum–terminal–velocity) of a bullet would be much less than 300 m/s. That means that it takes longer to fall to earth than it does to get to it’s zenith (peak).
Now, terminal velocity doesn’t effect vertical distance for a 90-degree shot. It just takes longer to fall. But on a 45-degree shot, once it reaches it’s zenith it doesn’t keep accelerating down until it hits the ground, terminal velocity would basically be like it gliding down after a time.
Think of a model rocket: it shoots up really really fast and then pops its parachute. Think of the parachute as air resistance. Once it parachute deploys, the rocket still accelerates downwards until the drag from the 'chute steadies it at a constant velocity and it glides to earth.
So considering only terminal velocity (affecting only vertical distance traveled), it will stay in the air longer and go further.
Factoring in air resistance’s horizontal component, even though it stays in the air longer, it’s always slowing down so it travels less far.
The question now becomes which magnitude is greater? I have no clue.
the deciding factor would be calculating how long gravity takes to slow a bullet to zero velocity vs how long it takes aerodynamic drag to slow a bullet to zero velocity. Zero velocity being the point where max distance has been travelled. The one with the longer time is the one that went the furthest distance. Its more complicated because drag is proportional to the speed, so as the speed slows the drag force gets lower, so it isnt a constant force. Gravity tho is always constant. I have the feeling you’ll have to use an integral to find the average drag force as it is not a constant force at all speeds.
I will guarantee you gravity will slow it faster than aerodynamic drag. You can reasonably assume gravity is already much stronger (in this particular situation) than aerodynamic drag since we are no where near terminal velocity of a bullet. Since gravity is much stronger, drag has little bearing on the vertical scenario.
If you want the angle to maximize the horizontal distance (assuming there is a ground and is infinitely flat) it is always 45 degrees. The 45 degree angle maximizes both flight time and horizontal speed to give you maximum horizontal distance travelled.
Okay, these answers are getting way too complicated. It is really simple, so don’t overthink it.
The constants are weight, muzzle velocity, gravity, and drag coefficient. Therefore we do not need to include them in this discussion.
A bullet traveling perpindicular to level will travel farther than one shot at an angle to maximize horizontal distance.I believe that angle is around 30 degrees.
A projectile moving only up only has to overcome gravity and decrerasing drag in its travel. Also all of the energy used to propel the bullet is consumed as velocity drops to zero. Then gravity overtakes inertia and the projectile falls all the way back to earth. This travel would be more than double in length to the optimum distance travel.
The projectile aimed for maximum distance would never come close to the travel distance of the vertical shot on even just the upwards leg. The bullet will impact the ground with forward momentum provided by initial thrust only.
The exact numbers don’t matter a damn. The theory is as true for rocket science and aviation in general as it is in firearms.
Drag coefficient and terminal velocity need to be taken into consideration (and behave differently) in the 90- and the 45-degree shots.
The question was vertical-shot (90 degrees) vs. optimal shot (which is 45 degrees). Not sure where 30 is coming from.
False. Which numbers you use for constants don’t matter a damn. But, as you can see, using numbers gets you an answer based somewhat in reality while pulling stuff outta thin air gets you an uneducated guess.
You’re right in that I don’t need the exact numbers that I used above. You’re absolutely wrong, however, when it comes to disregarding equations and basic physics and just making stuff up. I don’t mean to be rude, but what you said isn’t just wrong… it’s the opposite of true.
What differences are there in drag coefficients in regards to a 90-, 45- and 0- elevation shot? Same air right? Given perfect/same conditions (mainly no wind) for both shots, what can change the dynamics of drag or were you referring to terminal velocity? Terminal velocity won’t become a factor in the 0- shot as it will fall to the ground far earlier than it slowing to the terminal velocities at an angled shot.
Serious question, not hacking on you. You seem pretty damn smart, and I like to learn.
i guarantee you the vertical distance will be much less than the horizontal distance due to the fact that gravity will be much stronger than air resistance at these speeds. We are nowhere near terminal velocity of an freefalling object the size of a bullet, dont even bother think about TV.
No offense taken. Let me explain further.
Terminal velocity would only affect the vertical shot on its return to earth, not the horizontal shot. A propelled bullet would travel faster than terminal velocity. This only comes into play for time, not distance. The distance is determined by the altitude gained from the propellant.
The 30 degrees comes from the supposed optimal distance trajectory. not sure what the exact number is. I thought it was thirty. It has been many years since I heard it and I was surprised it was not 45 degrees myself. for argument sake we can say 45 degrees.
Atmospheric lapse rate is an advantage for the vertical bullet. for argument sake we can disregard the drag benefits for the vertical shot. Gravity works on both bullets equally. both bullets will use all of the energy not used to overcome drag to overcome gravity. The vertical shot will use its energy to overcome gravity until all forward velocity is lost. The bullet will then reverse course and fall to the earth. The forward projectile will continue to travel with forward velocity until it impacts the earth.
Equations are important. I am not going to dispute that. My facts are solid though. I did make one assumption that I can see as misleading. I ASSumed that we were only taking into account the hypotenuse, or the distance from shot fired to impact.
What?
If you shoot a bullet straight up in the air, it will stop, duhh. And, it will reach a maximum terminal velocity as it falls. That much is a given. It will not accelerate (during the fall) any more than it’s terminal velocity. Everything has a terminal velocity. Not everything will reach it because it takes a longer fall for some things to attain it, but a bullet fired from a gun, straight up, will most definitely get high enough to attain a terminal velocity on its return trip to Terra Firma. A bullet would need to fall about 8-10 seconds (depending on the density and shape) to attain its terminal velocity. That would be fairly easy considering how high a bullet would go in the first place giving it ample flight time on the return trip to attain terminal velocity according to several sources. (Just google terminal velocity of a bullet/marble, you could spend all day reading)
I read the OP as sending a bullet to a new destination (e.g.: downrange).
Hence I did not take into account the return of the bullet (being shot vertically) as the horizontal bullet would not be returning.
Therefore the horizontal bullet travels “farther”.
The drag coefficients should be the same. The difference is that you’ve got two totally different math problems DESPITE the fact that the air is the same, the bullet is the same, and gravity and muzzle velocity are the same.
(Again, I had high school physics and then a year of physics in college–and that was a few years ago, so this is just my simple understanding.)
With the vertical bullet, you have a 1-dimensional problem.
Muzzle velocity: UP
Gravity: DOWN
Drag: DOWN on the way up, UP on the way down.
With the arc’d bullet, you have a 2-dimensional problem.
Muzzle velocity: UP component and HORIZ component (the forces look like a triangle)
Gravity: DOWN
Drag: UP/DOWN component and HORIZ component (triangular forces, again)
Now, here is why it matters:
Calculations for how long it stays in the air are 1-dimensional. So for the arc problem you need to strip out the horizontal component of drag and muzzle velocity.
Calculations for how far it travels are 1-dimensional. So for the arc problem you need to strip out the vertical component of drag and muzzle velocity.
The same was true of the “frictionless vacuum” problem but it turned out that it came out the same anyways. So why is drag different? My GUESS (and it’s just a guess) is that terminal velocity throws a wrench in the works and makes it different. If that’s true, then you can’t ignore drag. If it’s not true, then someone smarter than me should probably step in with an explanation or a link describing where everything cancels everything else out.
My point is that no one has yet explained why terminal velocity is irrelevant (going to read pilotguy’s latest post when I get home from work to see if it explains it) and I’m not ready to toss it out without understanding why first.
Complication:
Got it. I thought you were assigning equal value (when you said difference) to drag and terminal velocity.
Terminal velocity should only play a roll in the vertical shot as the bullet would actually slow down past (and eventually stop) the terminal velocity because it’s going straight up. At 45 degrees, I seriously doubt that it’d lose enough energy to drop below the terminal velocity. I’m still searching for the link where I read that (with hefty explanation) but I’m at work so it’ll probably have to wait till I get home.