reason?

Dang all that typed out, only to come to an incorrect conclusion.

your S equations aren't correct, they should start with initial velocity multiplied by time, with initial velocity being zero.


take for instance this one:dimensional analysis shows the equation is incorrect. The first component g times t would leave you with velocity, not distance. it should start off with initial velocity times time, which is zero. H = -1/2*g*t^2.

I think you are misusing some variables here. Check out my post.

This is simply not the case. Why? Because the best drag model here is quadratic (proportional to the square of velocity). And if you look at the force diagram, and find Fy (the y component of the drag force) it'll depend on Vx as shown in post 100. I feel like this best communicated through a diagram, and I wish it was easier to make one on here.

For the record, there is no need to crunch numbers here at all. Specific numbers are not important- you can just compare the generic forces. That's why when saying in ideal conditions that the balls fall in the same time, we don't have to specify the mass of the ball, etc....we just reason through it by figuring out which forces are acting on the ball.

It's simple you autists, go throw a baseball hard horizontally, then have a friend drop a baseball from same height, the one thrown will stay above ground longer...I dint care about your silly equations....go outside and try idiots

i mean i'm a fukin idiot and even i learned this in high school physics......

Yes it depends on Vx, but Vx and the angle theta are interdependent bro, such that the magnitude of the drag force will ALWAYS be scaled so that the vertical component is the same with a horizontal velocity as it would be from free fall.

A high Vx means a higher drag, which will cause Vx to decrease more quickly and cause theta to increase more quickly. That relationship will always cause the drag force vector's y component to be the same as from free fall.

I don't know how many times I can say that. You saying 'it's not the case' doesn't make it so, and I actually put numbers to paper to confirm it.



Please prove to me how the statement I made is not the case, and I'll concede

You'd be surprised how much intuition can fail you. This is why even greats such as Aristotle thought, for example, that heavier objects fall faster than lighter objects. Obviously he was wrong.

Short answer: same time

Long answer: they hit the ground at basically the same time, but the cannonball fired horizontally will stay in the air a negligible amount of time longer due to the curvature of the earth. I don't think it would have any lift because it is not aerodynamically shaped, it's just a sphere

they would land at the same time since the earth is flat

Same time because only thing that matters is vertical velocity and acceleration.

Cannon fired horizontally has the same initial vertical velocity (which is 0m/s) as the object dropped from the same height and the only force acting on them to accelerate them is gravitational force which acts vertically. Therefore with same initial vertical velocity, same vertical distance they need to travel (height)and same net force acting on them; the time is the same .

This is assuming theres no air resistance and chit tho. Basic first year physics

The error here is the components don't all scale each other out as you are saying. I'll explain this:

Say the drag forcewaslinearly dependent on v, the magnitude of the velocity of the projectile. Then we have F_drag = kv, for some constant k. Then, when we try to find the y component of said force, we would have

Fy = k*v*sinθ
Fy = k*sqrt(Vx^2+Vy^2)*sinθ
Fy = k *sqrt(Vx^2+Vy^2)*(Vy/sqrt(Vx^2+Vy^2)

Fy = k*Vy

(Note the square roots cancel and here we have a situation where the vertical component of drag force IS independent of Vx....it just depends on Vy and will be directly analogous to the vertically dropped ball).

The problem is, when drag from air is quadratic (and as noted above, it is best modeled this way because it is a cannonball through air), you get a different result:

Fy = k*sqrt(Vx^2+Vy^2)*Vy

Which varies with Vx (and no, the relation to sinθ is not important at this point, that was already factored in by writing sinθ= Vy/sqrt(Vx^2+Vy^2)

I stopped reading after page 1. Imagine doing undergrad in physics and finishing grad school, then reading this thread. Do it in that order and be thankful you were a STEMcel. People's ignorance abouteverythingwill annoy you at first but you get used to it and humble yourself.

People don't know basic kinematics. It isn't surprising at all, its more surprising they're so bad at basic math. Math is easy. Setting up a real problem where you arrive at derivations or equations with a real, working system in industry is not.

Can't believe now 6 page thread
This is high school mechanics
Good to see physics on the misc though

Calm down bro. I'm sure there are a myriad of topics that you don't know chit about and other people are experts on. Even if someone took AP physics in high school, if you aren't consistently using the material, you are going to brain dump it. When I got out of the military and took my math placement test at a community college, I placed in a remedial math course because it's been four years since I've seen trig or factored anything or seen a log.

Maybe the math associated with basic high school physics, as discussed in this thread (I wouldn't even call that math, I'd call it calculation).

I could load this post up with partial differential equations arising from heat flow or fluid dynamics, and ask to solve them with techniques from Fourier analysis.....then.....the math is not so easy eh.

Of course also there are problems, typically regarding more abstract subjects such as number theory, that make the above problems about PDEs look like pieces of cake.

True! But for some reason, all of Dynamics is with me all the time even though I don't use it. There are things just burned into my brain. The more school I did the more I realized how little I knew.Probably. The 'monkey in a tree' problem is as iconic in a 1st semester physics class as the 'slipping ladder' problem in basic calculus or the 'tipping problem' in mechanics. I just assume too much is all.

That feel when you post in a physics/math thread with absolutely no knowledge other than what you remember from high school and no one calls you out

Feelzgoodmayne

well, you dont need to be a phd in theoretical physics to figure this one out lol..

we literally proved this mathematically in like grade 10...

My CP looking like a warzone now... stay rustled *******s.

The question of a fired vs a falling bullet is an interesting thought experiment. It would seem to me that a bullet fired with significant velocity at a perfectly horizontal angle to the immediate plane of the earth (in other words perfectly perpendicular to the earth's gravitational force) would hit the ground after the other bullet, being of identical size and mass, which is dropped from the same exact height above ground. this even assumes that the test is conducted over a topological flat ground. Bullets falling to the ground, whether by dropping or firing, are set into motion on a linear path toward the gravitational center of the earth. In this example we can ignore the slight gravitational variations the fired bullet would travel since the increase of gravitational force would be too small to appreciably affect the result. The main issue here is that by firing the bullet we have applied energy to the bullet that has acted on it in contradiction to its original linear path; we have set it on a new linear path. The two will begin to average out, forming a parabola (or bullet drop) over distance, as the two forces work against each other. But, since we have applied energy to the mass in a linear direction different than that of the linear direction of the gravitational energy, we have acted on it, which as Newton so aptly pointed out, will affect its tendency to "stay in motion". The problem with the thought experiment is that of scale. If you step back and look at it from a larger scale, like that of the bullet's relationship to the entire planet (and the spherical curvature of space-time surrounding our planet's gravitational center) it makes more sense. When you fire that bullet, you've essentially set that bullet in motion against that curvature. Satellites are a good example. Ignoring the concepts of "geostationary" as that is not relevant here, this should make sense. The way that a satellite stays in orbit is by getting it up to the desired height and applying a linear momentum to the satellite perpendicular to the earth's gravitational field. In this way, the satellite will want to travel in a straight line, conserving linear momentum, but the earth's gravity will "tether" the satellite as it exerts energy to pull the satellite along its linear path. The result is the two forces fighting each other: The satellite can't fly off into deep space because of the gravitational "tether", and the satellite won't just fall straight to the earth due to its desire to travel in a linear path perpendicular to this other force. If you balance these forces, the satellite maintains orbit. If, however, you launch the same satellite up to the same height and fail to give it any linear momentum contradicting the gravitational attraction of the two bodies, the satellite will simply fall back to earth, following its linear path along the curvature of space-time towards the gravitational center. The bullets are the same way, only on a much smaller scale. Theoretically, if you fired the bullet with sufficient energy as to have equal energy in both perpendicular paths, take out obstacles and drag coefficient, and could theoretically maintain the linear energy by adding additional energy to the fired bullet as it travels, it would essentially orbit the earth at 3 feet from the ground. It's a matter of energy, of which mass is only a portion of the equation. Velocity is the other portion of that equation. I can imagine that if the test were conducted with the use of a 50 caliber bullet or a large rail gun then the time differences would be such that it would be easier to see the differences between a dropping a bullet with no energy or momentum in the perpendicular linear path or dropping a bullet with vast quantities of energy and high velocities in the perpendicular linear path. I'm by no means a physics expert or professor so if my terminology is off then I apologize, but the concept is evident. Perhaps this angle (pun partially intended) is the one that has been overlooked in the thought experiment because it's hard to imagine a tiny bullet in scale to the entire earth and the space-time in which the earth sits and is traveling.

go back to my post about lift. It is absolutely possible being a sphere, I mean have you never heard of a curve ball? But like I said previously, it would have to be massive and deliberate.

Holy **** I truly hope that most of you on page 1 were trolling. "But the velocity it's fired out of the cannon at will dictate how long it stays in the air". No, it will dictate how far it travels on a horizontal axis. On the vertical axis it is subjected to the exactly same 9.81 m/s^2 that the one being dropped is.


Edit: typos

This is retarded as fuk

Gonna start a go fund me to do the experiment myself

Age: 6
mirin

This is actually a really great thread because it shows how important certain assumptions are, and how it is always important whenever you model anything to carefully state your assumptions and keep them in mind when you explain your result. The "same time" answer is the textbook answer, and it holds assuming the Earth is flat, the ball has a relatively low velocity compared to the escape velocity of the Earth, there is no air resistance, there is no spin on the ball, we assume the ball is a point, the mass of the ball is small relative to the Earth (and hence the acceleration due to gravity is constant), and probably a few other assumptions I have forgotten to say. The most realistic assumption of these to change is that of air resistance, in which case, we can see that the balls actually do have different travel times depending on initial horizontal velocity, as seen from numberguy's argument. It is also interesting that quadratic drag motion in 2 dimensions cannot be solved analytically and must be solved numerically. This shows how even simple problems that everyone is familiar with can turn out to be mathematically quite difficult to deal with exactly.

Interestingly, many people who play a lot of sports had the right intuition and came to the right answer without any scientific reasoning lol. It sort of demonstrates why we evolved to actually be so bad at math. Many of the calculations required for our daily life would be intractably hard to solve analytically, so we are only ever always using approximations and analogies to numerical techniques and numerical optimization schemes to find "good solutions" to the problem rather than the exactly correct one.

[QUOTE=AquilaConfido;1549202931]2:1 Mathematics degree here, and you're wrong. Speed matters when cannonball is fired horizontally, because F=ma.

When cannonball is dropped, equation becomes F=mg because gravity.

When fired from horizontal, acceleration must be taken into account because that will have an effect on how long the projectile will travel before it hits the ground.

Basic projectile ballistics.[/QUOTE

You must have failed physics

Same time unless dropped on an edge of a tornado . Same goes for bullets

This is an interesting point. Didn't realize it at first, but the system of differential equations representing the motion equations for each component do indeed look like they are not solvable analytically.

In a physics class textbook, they hit at the same time cuz no wind resistance, no friction, or other external forces other than cannon ball being fired perfectly horizontal and the other cannon ball being dropped at the exact same time... In real tea, the cannon ball being dropped would hit first...