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Forum » A cannon ball is fired from a cannon horizontally
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Thread: A cannon ball is fired from a cannon horizontally

post 1549225491 04-16-2018, 10:40 AM
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Lol @ the pathetic kunts who negged me about chit I studied 15+ years ago.

Came to the UK permanently when I was 14, English isn't even my first language, started university at 17, finished at 20 years old from top 20 UK university with a Bachelor in Maths.

IQ around 130 mark.

U mad?

post 1549225691 04-16-2018, 10:42 AM
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At what yardage does the curvature of the Earth become apparent?

Inb4 millenials
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post 1549226241 04-16-2018, 10:48 AM
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Originally Posted By AquilaConfido⏩
Wrong. The faster the initial speed of the projectile, the longer it will travel and the more time it will take to hit the ground.
THis and you have to count in the curvature of the earth that gives it some more room to fall.

















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post 1549226461 04-16-2018, 10:51 AM
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Yellow
post 1549226481 04-16-2018, 10:51 AM
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Originally Posted By MakeABanana⏩
Ya'll should look into how satellites stay in orbit.
This. It’s not rocket scienc OP
post 1549226641 04-16-2018, 10:53 AM
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Originally Posted By AquilaConfido⏩
Lol @ the pathetic kunts who negged me about chit I studied 15+ years ago.

Came to the UK permanently when I was 14, English isn't even my first language, started university at 17, finished at 20 years old from top 20 UK university with a Bachelor in Maths.

IQ around 130 mark.

U mad?
>130 IQ
>can't understand motion in x and y planes don't affect each other


nah bro I'm not mad, kind of feel bad for you. Most miscers that answered correctly ITT probably don't have an engineering or physics degree, they're just smart enough to understand this concept at face value.

And I negged to dissuade you from chiming in on things you don't understand, btw. Don't blame the fact that you don't understand on how long it's been since you got your math degree.
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post 1549226761 04-16-2018, 10:55 AM
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Originally Posted By Jacobcapra⏩
Not sure if srs, hope this is trolling.


mechanical engineering degree checking in and you're both potatoes.


F=ma is true, but acceleration in the x-direction will not affect acceleration in the y-direction. For both cannonballs, the only force acting in the y-direction is gravity. They will have the same acceleration in the y-direction, and will land at the same time (excluding the difference in distance dropped due to the curvature of the earth).


edit: I see you have already been shown that. Also cringing so hard at the fact that you were being srs.



You don't even have to disregard air resistance. The air resistance in the Y-direction will be the same for both cannonballs
Not correct. Vertical component of drag will vary with horizontal velocity.

Consider the horizontally fired projectile at a certain point during its trip. The drag will be in the direction opposite of its velocity vector with magnitude K(Vx^2+Vy^2), where K is a constant and Vx and Vy are the component velocities of the projectile. The magnitude of vertical component of drag will then be K(Vx^2+Vy^2)sinθ, where θ is the angle the projectile's trajectory makes with the horizontal (sinθ = Vy/sqrt(Vx^2+Vy^2)). Clearly Vx is affecting vertical component of drag.......and Vx will be higher for the shot cannonball (compared to Vx= 0 for the dropped).

Sure, depending on the situation, the effect of drag will be too minuscule to measure, but that doesn't change the fact the length of time the projectile takes to hit the ground does depend on air resistance (as well as curvature of earth, wind, etc)

I would think a "mechanical engineering degree" would know this, it's pretty elementary physics. Lol at your comments about math majors. Hopefully you aren't building bridges.
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post 1549226851 04-16-2018, 10:56 AM
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Originally Posted By Kylevolution⏩
Made a diagram for you *******s who are struggling,



If the velocity of the cannonball exceeds perpendicular gravitational pull then it will
WOW, did you pass middle school?
<HTC>
post 1549227041 04-16-2018, 10:58 AM
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same time. That is basic high school physics
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post 1549227111 04-16-2018, 10:59 AM
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Originally Posted By AquilaConfido⏩
Lol @ the pathetic kunts who negged me about chit I studied 15+ years ago.

Came to the UK permanently when I was 14, English isn't even my first language, started university at 17, finished at 20 years old from top 20 UK university with a Bachelor in Maths.

IQ around 130 mark.

U mad?

Mechanics =/= maths

We arent even calculating anything *******
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post 1549227251 04-16-2018, 11:01 AM
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#101
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Originally Posted By Jacobcapra⏩
>130 IQ
>can't understand motion in x and y planes don't affect each other


nah bro I'm not mad, kind of feel bad for you. Most miscers that answered correctly ITT probably don't have an engineering or physics degree, they're just smart enough to understand this concept at face value.

And I negged to dissuade you from chiming in on things you don't understand, btw. Don't blame the fact that you don't understand on how long it's been since you got your math degree.
You sound like some Redditor loser who hopped on OJ and now thinks he's king of the world.
Originally Posted By LukeS1⏩
Mechanics =/= maths

We arent even calculating anything *******
Location: Brighton.
post 1549227291 04-16-2018, 11:02 AM
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#102
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Originally Posted By AquilaConfido⏩
I can confirm, I was second neg from the top.
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post 1549227691 04-16-2018, 11:07 AM
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#103
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Originally Posted By Jacobcapra⏩
Actually, I really don't think they do.





Pretty sure engineers build bridges, not math majors


lol @ miscers negging back cause they're butthurt. You earned your neg by pontificating about a topic that you don't understand, can't have armchair engineers spouting off garbage. Truth hurts, brahs
This is hilarious coming from someone who doesn't even understand basic air resistance and vectors.
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post 1549228011 04-16-2018, 11:10 AM
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Originally Posted By LukeS1⏩
the one you drop, the one from the cannon travels laterally so due to the spherical nature of the earth will have further the drop (assuming its on "flat" ground)
jokes on you everyone knows the earth is flat. right?
You can always go deeper than you think - Chris Bumstead
post 1549228761 04-16-2018, 11:18 AM
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#105
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Originally Posted By AquilaConfido⏩
You sound like some Redditor loser who hopped on OJ and now thinks he's king of the world.



Location: Brighton.
You are literally a 34 year old virgin using my location as insult, which actually happens to be one of the highest quality of life cities in Europe. Take a look at yourself *******
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post 1549228821 04-16-2018, 11:19 AM
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#106
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the cannonball came to the puddle, OP said how many were in the puddle
post 1549228851 04-16-2018, 11:19 AM
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#107
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Originally Posted By numberguy12⏩
Not correct. Vertical component of drag will vary with horizontal velocity.

Consider the horizontally fired projectile at a certain point during its trip. The drag will be in the direction opposite of its velocity vector with magnitude K(Vx^2+Vy^2), where K is a constant and Vx and Vy are the component velocities of the projectile. The magnitude of vertical component of drag will then be K(Vx^2+Vy^2)sinθ, where θ is the angle the projectile's trajectory makes with the horizontal (sinθ = Vy/sqrt(Vx^2+Vy^2)). Clearly Vx is affecting vertical component of drag.......and Vx will be higher for the shot cannonball (compared to Vx= 0 for the dropped).

Sure, depending on the situation, the effect of drag will be too minuscule to measure, but that doesn't change the fact the length of time the projectile takes to hit the ground does depend on air resistance (as well as curvature of earth, wind, etc)

I would think a "mechanical engineering degree" would know this, it's pretty elementary physics. Lol at your comments about math majors. Hopefully you aren't building bridges.
Way to overcomplicate things, and not even be correct, brah.

Vy will be the same for both the dropped and the shot ball, sothe vertical componentof air resistance must be the same for both.

The relationship between air resistance, Vy and Vx will always be such that K*Vy*sin(90) = K(Vx^2+Vy^2)*Sin(theta)


who was dependent variables. Wish I could find a good online model and prove this to you
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post 1549229101 04-16-2018, 11:23 AM
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#108
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Anyone else thinks that the big bang was caused by something outside of time and space?
Everything I post is trash.
post 1549229511 04-16-2018, 11:28 AM
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#109
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Originally Posted By AquilaConfido⏩
You sound like some Redditor loser who hopped on OJ and now thinks he's king of the world.
Check the user title, brah. we don't drink the OJ.
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post 1549229801 04-16-2018, 11:30 AM
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#110
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Originally Posted By LukeS1⏩
You are literally a 34 year old virgin using my location as insult, which actually happens to be one of the highest quality of life cities in Europe. Take a look at yourself *******
Brighton - gay capital of the UK.
Originally Posted By Jacobcapra⏩
Check the user title, brah. we don't drink the OJ.
Yeah brah, I believe you, and pigs fly.
Originally Posted By peterplsss⏩
I can confirm, I was second neg from the top.
Fukking loser.
post 1549229931 04-16-2018, 11:32 AM
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#111
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Originally Posted By Jacobcapra⏩
Way to overcomplicate things, and not even be correct, brah. Also you missed a square root in your vector equation. K*Sqrt(Vx^2+Vy^2)


Vx will be the same for both the dropped and the shot ball, sothe vertical componentof air resistance must be the same for both.

The relationship between air resistance, Vy and Vx will always be such that K*Vx*sin(90) = K(sqrt(Vx^2+Vy^2))*Sin(theta)


who was dependent variables. Wish I could find a good online model and prove this to you
Again, nope. You're digging a deeper hole here.

Air resistance F will be kv^2.....

v = sqrt(Vx^2+Vy^2). (It's the hypotenuse of a right triangle with Vx and Vy as legs)

So F = kv^2 = k(Vx^2+Vy^2), no square roots.

As far as your second paragraph, this makes no sense whatsoever. Vx is the same? Horizontal velocity is the same? For shot and dropped cannonball? If you mean Vy....this is still not correctwithair resistance, as outlined above. The vertical component of air resistance is affected by horizontal velocity.

Also, this is not overcomplicating things. This is 10th grade physics.
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post 1549230121 04-16-2018, 11:33 AM
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#112
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Originally Posted By AquilaConfido⏩
Brighton - gay capital of the UK.



Yeah brah, I believe you, and pigs fly.



Fukking loser.
If you are going to troll at least put effort into it. No respect for the art of trolling
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post 1549230311 04-16-2018, 11:35 AM
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They hit the ground at the same time.
U
post 1549230441 04-16-2018, 11:36 AM
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#114
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Lmao @ this thread
post 1549230641 04-16-2018, 11:38 AM
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#115
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Originally Posted By axiom15⏩
Lmao @ this thread
Just reading it now. Started off with a physics question and turned into a WWE royal rumble lol. Misc please never change.The answer is nothing happens. How can gravity be real if weight isn't real? *mind blow*
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post 1549230751 04-16-2018, 11:39 AM
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Originally Posted By LukeS1⏩
If you are going to troll at least put effort into it. No respect for the art of trolling
Go away plz.
post 1549230761 04-16-2018, 11:39 AM
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Originally Posted By AquilaConfido⏩
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.
You're wrong. We actually did this experiment in class.

They hit at the same time if fired with 0 elevation.
U
post 1549231041 04-16-2018, 11:42 AM
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#118
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Originally Posted By numberguy12⏩
Again, nope. You're digging a deeper hole here.

Air resistance F will be kv^2.....

v = sqrt(Vx^2+Vy^2). (It's the hypotenuse of a right triangle with Vx and Vy as legs)

So F = kv^2 = k(Vx^2+Vy^2), no square roots.

As far as your second paragraph, this makes no sense whatsoever. Vx is the same? Horizontal velocity is the same? For shot and dropped cannonball? If you mean Vy....this is still not correctwithair resistance, as outlined above. The vertical component of air resistance is affected by horizontal velocity.
yes I meant Vy, also realized you were taking the drag coefficient and density in the drag force equation and calling it a constant K, not calculating a resultant vector from the vertical and horizantal velocity components which would introduce the square root.


That being said, it is not true that horizontal velocity affects vertical drag.... otherwise the hundreds of experiments exactly like the OP would not have found the exact opposite conclusion.

Contemplating actually crunching the numbers to prove it
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post 1549231131 04-16-2018, 11:44 AM
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#119
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Alright. This is ridiculous. Let me show you *******s how you do this so you can all stfu. Let's assume there is no air resistance. Let's assume the cannonball is at a height h above the ground at the initial time t=0. We will set the origin at the initial position of the ball. Thus, the initial position of the ball will be (0,0), and the final position of the ball will have y coordinate -h. The ball will have initial horizontal velocity v0x, and initial vertical velocity v0y. The acceleration due to gravity of the ball will be assumed to be constant with a magnitude of g and a sign of -1 due to our coordinate system, the acceleration horizontally will be 0 since there is no rocket booster or any other such chit on the ball, and we will treat the ball as a point particle. Air resistance will be neglected. We will assume the Earth is approximately flat and the velocity of the cannonball is much lower than the escape velocity of the Earth. Then, the equations of motion are simply given as follows:

x_f=v0xt+0
-h=v0yt-(1/2)gt^2

Now note that t is the same for the system of equations since it is the travel time to get from the initial to the final position for the ball. Now, we note that if the ball is fired horizontally, it has no y component for its initial velocity so v0y=0. Then, if we write the y equations for each situation, we have 2h=gt^2 for the cannonball if fired horizontally and 2h=gt^2 for the cannonball if dropped from rest. Notice that the mass (and therefore, the weight) of the balls is not relevant at all under these approximations. (It would only become relevant if it were large enough relative to the mass of the Earth.) The ball takes approximately sqrt(2h/g) units of time to hit the ground.
post 1549231311 04-16-2018, 11:45 AM
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#120
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If the canonball is shot with enough velocity it could theoretically escape earth's gravity even if fired horizontally due to the earth being round
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