I've been trying to understand Newtonian mechanics and kinematics for days now. I didn't know where to start so I thought I'd start from a scenario where Tormund drops a rock from the Wall and I calculate how long it takes to hit the ground.
It took me 2 days to integrate these equations starting from only the fact that the downwards acceleration is -g. Did I get these equations right? I still don't know how to deal with the minus, for example.
Next up I will tackle my greatest enemy: projectile motion. Let's say Ygritte fires an arrow from ground to the top of the Wall (210 meters). I'll try to calculate the initial velocity required.
Just set downward motion as your positive axis dumbass...
Zachary Martinez
Why is it -g in the first place then
Ryder Adams
Because conventionally up is positive and down is negative. Gravity points down so g is typically seen as negative. If you were to say down is positive and up is negative then you have a +g. You can orient the direction of your axis in any way so long as you follow the "right-hand rule." Google that if you don't know what that is.
t. Fizix mayger.
Joseph Sanchez
Why is the speed of light always constant no matter who observes it?
Isaac Allen
Idk. Something to do with Maxwell's equations and Einstein being really smart.
William Evans
Well the mass is irrelevant so you go s(t)=1/2*a*t^2 Solve for t Squr(2*s/a)=t Put in what you have It should be like 6.5 seconds or something
Ayden Martinez
>Einstein being really smart
Was Einstein really that smart or is it a meme? I see shit nobody else sees, too. Like I'm pretty sure there's a debug mode in the universe where time doesn't exist and sometimes I live in that dimension all alone.
Ayden Miller
Those As you crossed out in your second integration represent your initial x and y positions. Ignore the x but the y is where you will plug in your initial height. Set the whole equation to zero and solve.
Brandon Scott
A question about projectile motion..
Why isn't the greatest distance for a projectile achieved when the initial angle is as close to 90 degrees as possible? Like, if the angle is close to zero, gravity will pull down your arrow (or whatever) and it will hit the ground. At 90 degrees it will just go up and come back down. But what about 89 degrees? It should be optimal, but I've heard 45 degrees is the answer.
Blake Long
For initial velocity you can go like this: Since you know that the rock takes 6.5 seconds to the ground, the velocity will be the same the arrow will have to have because with the same g it will take the same time to decellerate from some velocity as to accelerate to it. So you go v(t)=at Set a=9.81 and t=6.54 You get like 65 m/s
Oliver Ortiz
Intuitively: Because the "sideways" and the "upwards" part of the motion will have the same value Mathematically: Set in equation for diagonal motion, integrate, set for zero and solve the answer will be 45 degrees
Samuel Bailey
Uh doesn't the shooting angle matter a great deal?
Zachary Sanchez
I think you might be the schizo retard.
Jace Fisher
>Because the "sideways" and the "upwards" part of the motion will have the same value
I don't even intuitively understand how having the same value in both helps the arrow go far In video games you always make your grenade go the furthest by throwing it as fucking high up as possible
Jack Evans
The mass isn't irrelevant when wind resistance exists But usually we disregard it
Jonathan Wood
>I think you might be the schizo retard.
I'm clinically insane, yes. And I have no concept of ordinary logic like normal people do. But I am a physics student in a university. I honestly believe I'm going to be great because of how much I can dip in to the insane side and get secret equations from there, far from the reach of normies.
Landon Powell
>In video games you always make your grenade go the furthest by throwing it as fucking high up as possible No you dont. It will only have the longest time in air if you do that
Cameron Butler
Why does mass help shit fall faster? Does it somehow push air molecules outta the way better?
Jace Cook
Long airtyme will have the farthest reach because x = v * t
Noah Flores
I will never get a gf because my initial epicness vector has no acceleration in the social direction and stays the same through life. My e_s initial = e_s final I only have acceleration in the autism direction
Second law of newton, the heavier something is the lesser will be the acceleration applied on it by other things An object in free fall is constantly hitting air particles on it's way down, which are applying a force contrary to the object's direction, dampening the fall. The heavier the object is, the lesser will be the acceleration caused by those forces Now the wider something is, the more air particles will be resisting it's fall. That's why parachutes need to be light and wide
Carter Martin
idk but i found learning them by thinking of mathematics and not physics was easier. There's some "physics for mathematicians" books out there that i recommend if you like that route
Brandon Garcia
That is wrong. That equation only works for movement in one direction
Andrew Adams
>mfw teaching high schoolers basic physics on Jow Forums seems like i'm putting my degree to good use
Noah Price
Do not draw your sword to kill a fly. Integration is absolutely not needed in these simple problems. Anyways, downwards acceleration is g. It is positive or negative based only on your choice of axis directions. There is no need to use the right hand rule suggested as you are only dealing with 2D movement, and as such need no z axis. Also, you seem to have done it right it your attempt, by making the distance(210m) and the acceleration(g) one be positive and the other negative. Note that the order does not matter, as the axis is arbitrary. The only mistakes you have made are using the lesser and greater symbols to denote the vector(that is the notation for internal product, use parenthesis instead) and you have taken the square root without care - if t squared is 2(210)/g, t is plus or minus sqrt(420/g). As you've set the initial time to 0, the correct answer is the positive one.
Jose Lopez
Partly true. However, that v is velocity in the direciton of x, which depends on the angle. For 90 degrees, it is 0, giving the least range out of any angles, tied with 0 degrees.
Considering a flat surface, the projectile will arive at it's destination in time T, with vertical velocity opposite to what it started, as the movement is symetrical. Let vy be the vertical velocity and g the acceleration(which is in the opposite direction, so if one is positive the other is negative): velocity=starting velocity+ acceleration multiplied by time => -vy=vy+(-g)T => T=2vy/g. With v the initial velocity, T=2v(sin(angle))/g. Range will be horizontal velocity which is v(cos(angle)) multiplied by the time it is moving, so range=v(cos(angle))2v(sin(angle))/g. Using the trigonometry identity sin(2(angle))=2cos(angle)sin(angle), we simplify it to range=(v^2)sin(2(angle))/g. The angle is between 0 and 90 degrees, so sin(2(angle)) is between 0 and 1. To get the maximum distance, sin(2(angle)) must be the maximum value (1). sin(90deg)=1 => 2(angle)=90 => angle=45 degrees for maximum distance. You can do this searching for maximum values using calculus but that's wack and slower.
Noah White
That is a weird question. You probably know about refraction, so I assume you mean the speed of light in a vaccum. Anyways, we think that is true because it is testable but more importantly because it makes good predicitons. As in, special relativity is a thing if we consider that to be true, and special relativity is testable and has made accurate predictions of the state of things.
Ryan White
You drop a rock into a well. 3 seconds later, you hear the splash. How deep is the well?
Liam Harris
i actually love tutoring heh. if you want a kinematics tutor hmu sweet knees#5848
Joshua Collins
9.81 meters times 3?
Brayden Hernandez
no origgz
Thomas Perry
No, that'd tell you what its final velocity is.
Robert Wood
During the first second rock travels at 9.8m/s, during second second it's 19.7m/s, during thurd second it's even more..
So the well is almost 60 meters deep? The hell
Bentley Miller
I was learning Inverse Kinematics for a robotics moocs and I ended getting to exhausted to continue after trying to understand DH parameters.
Carson Morgan
So I have to use that 0.5gt^2 equation? Is it possible to intuitively find the answer?
David Robinson
0.5gt^2 is the integral of g*t integrated by dt, so that equation is intuative.
Noah Garcia
You need to consider that, before you hear the splash, the sound of the rock hitting the water needs to travel all they way back up to you The speed of sound on air is 343 m/s.
Jayden Nelson
itt : pretentiouss retards
Lincoln Clark
You shouldn't need to remove the minus to make it work. Imagine up is postive and down is negative. Imagine where we are as the origin, a.k.a. we are at 0 on the y-axis. if the rock drops and its accelerating down (towards the earth), it's negative. When it hits the ground its traveled 210 meters from it's original point. That's our displacement -210 meters.
Since we have -9.81m/s^2 and -210 m, and doing a bit of algebra the negatives should cancel out
Matthew Butler
>gcse physics is pretentious
Carson King
The attempts at answering its pathetic
Carson Anderson
It's literally all just basic maths
Brayden Phillips
Ignoring perception speed and air resistance, considering gravity g and the speed of sound Vs: Let t1 be the time the rocks hits the bottom, t2 be the time the sound takes to reach the surface, D the depth of the well and Vs the sound speed. (g(t1)^2)/2=D, D=Vs(t2) => t2=g(t1^2)/2Vs t1+t2=3 => (t1)+g(t1^2)/Vs=3 => t1=((-1(+ or - here but it's + since time is positive)sqrt(1+6g/Vs)/(2g/2Vs)) => t1=(-Vs+sqrt(Vs^2 + 6gVs))/g D=g(t1^2)/2=((sqrt(Vs^2+6gVs)-Vs)^2)/2g=>D=(Vs^2)/g - Vs(sqrt(6Vsg+Vs^2))/g + 3Vs took too long
Charles Robinson
>During the first second rock travels at 9.8m/s, that's not correct. if t's accelerating at 9.8m/s2 and started with 0 speed, it will only hit 9.8m/s at the end of the second. at 0.1 seconds it will be travelling at 0.98 m/s, 4.9 m/s at half of the first second and etcetera. You need to use the formula
Landon Rodriguez
Some were. I like to think mine were pretty good: Do you disagree?
should we create a group? anyway, if someone wants help drop contact info or just ask here
Nathan Sanders
I now realize you were calculating the rocks descent, not using a clever way to calculate the arrow speed. Then you should've had both 210m and g positive or both negative, as they have the same direction.
Evan Foster
Dump incoming, so ye hope this will help, I just tried to do the question the way I would, hopefully it's right
