Talk:Trajectory of a projectile

message: Thanks to whoever made a cleaner  of my free body diagram. Looks great, except that those x and y indices on v and F(sub a) are subscripts, not exponents. I'd really be grateful if you could just touch it up for me. Thanks. --Random nick 05:16, 5 February 2006 (UTC)
replies: text: The free body diagram is excellent, but it does not print properly.  It shows up as a black square when printed.  I have seen this problem with images captured from PowerPoint as jpg before.  If a .tif format is used, this problem goes away.  Thanks, 65.165.72.203 19:00, 24 July 2006 (UTC)
thread_title:  The free body diagram 

message: It's a high school problem... solved... (No trolling intended... just being clear) --euyyn 22:54, 25 July 2006 (UTC)
thread_title:  This article doesn't belong in Wikipedia<span id="This_is_not_an_encyclopedia_article"></span> 

message: I tagged the  section with  instead of . The section doesn't require cleanup in terms of the Manual of Style, but it does have a series of equations that to a person unfamiliar with topic (e.g. me) doesn't seem to make much sense and/or offers no explanation as to what is occurring. They seem to be just thrown in together. Barkeep Chat | $ 13:30, 10 June 2008 (UTC)
thread_title:  Tagged with {{tlf|expert}} regarding equations<span id="Equations_-_Expert"></span> 

message: 9.81 ms-2? Why is the exponent -2? Shouldn't that be 9.81 ms2? I'll fix. — Preceding unsigned comment added by Petercorless (talk • contribs) 20:49, 19 February 2009 (UTC)
replies: text: You won't. Meters per second, per second acceleration is indeed ms-2. (User) Wolfkeeper (Talk) 20:59, 19 February 2009 (UTC)
replies: text: Of course ms2 would be wrong, but m/s2, as it has now been changed to, is just as correct as ms-2 and much more readily understood. —JAO • T • C 22:31, 19 February 2009 (UTC)
thread_title:  Negative exponent?<span id="Odd_Gravity.3F"></span> 

message: I propose to remove from the  section the lengthy explanation of why k has the units of N*s/m. It is obvious from Fair = k*V (if Fair has the units N and V has the units m/s then k must be N*s/m). My opinion is that the extra explanation is overkill. Skorkmaz (talk) 17:46, 29 October 2009 (UTC)
thread_title:  Remove lengthy explanation 

message: It would seem that in the air resistance section it puts g at -9.8 ms-2, shouldn't it be positive? I checked it out (in Garry's Mod of all things) and it came out with a negative value for an angle when calculating the angle of reach. 219.89.205.8 (talk) 05:24, 28 November 2009 (UTC)
replies: text: The sign of g will depend on what sign you take your velocity vector on. If you fire a gun upwards at a positive speed, it will have an acceleration of -9.8 ms-2. If you fire it downwards at a positive speed, it has acceleration, g, of 9.8 ms-2. Hopefully I am not being patronising, but the sign depends on what you have taken to be the positive direction. 129.234.252.67 (talk) 16:41, 11 February 2010 (UTC)
thread_title:  Negative gravity?<span id="Gravity_again"></span> 

message: Firstly, this article only deals with when drag is proportional to speed, which isn't so since for faster speeds F=cv^2. Secondly, the article doesn't deal with how you account for the curvature of the Earth and Coriolis force and centrifugal force. I could add how to deal with F=cv^2, but curvature of the earth depends on where you are, same thing with Coriolis force and centrifugal force. I could add how one accounts for that, but it's going to take some time, maybe during the spring break. So if someone else wants to make additions, please feel free. Pseudoanonymous (talk) 22:46, 17 January 2010 (UTC)
thread_title:  Things to add to the article<span id="Further_additions"></span> 

message: Maybe  isn't correct. I believe that a ballistic re-entry trajectory for a spacecraft does include drag, but not lift or propulsion. Terminology for artillery projectiles may be different. Martijn Meijering (talk) 22:34, 15 February 2011 (UTC)
thread_title:  Ballistic spacecraft re-entry<span id="Ballistic_trajectory"></span> 

message: Is the max. height always 1/4 of the range when the angle is 45 degrees and air friction is being ignored? --SamForestell (talk) 22:32, 20 November 2011 (UTC)
replies: text: Yes, providing the maximum height and range are very small compared with the radius of the Earth. DOwenWilliams (talk) 00:55, 15 March 2013 (UTC)
thread_title:  Question about height/range at 45 degrees<span id="Question_about_height.2Frange_ratio"></span> 

message: the animation claims a "max angle" at 32 deg in a vacuum.  But there is no x acceleration so the projectile will continue to increase indefinitely at the same rate, irrespective of initial angle.  — Preceding unsigned comment added by 149.241.5.104 (talk) 04:09, 26 March 2013 (UTC)
thread_title:  Animated GIF in section [[Trajectory of a projectile#Angle required to hit coordinate (x,y)|''Angle required to hit coordinate (x,y)'']] 

message: In both the animation and the diagram near the top of the article, it is stated that the launch speed (not x-velocity) is constant for all launch angles. By Pythagoras:
:Vl=\sqrt{Vx^2+Vy^2}
Since Vl is constant, the square-root must be constant, so the larger Vy is (steeper launch angle) the smaller Vx is. When the launch angle is small, Vy is small, so the projectile quickly drops to the ground. It doesn't travel far even though Vx is large. As the angle increases, Vy increases, so the projectile remains in flight for longer and travels further. But when the launch angle is large (more than 45 degrees if the heights of launch and arrival are the same), Vy is large, so the projectile remains in flight for a long time, but Vx is small so it doesn't go very far. In the final case where the launch angle is 90 degrees, Vx is zero, so the projectile goes straight up, then straight down, landing where it started. DOwenWilliams (talk) 15:08, 26 March 2013 (UTC)
thread_title:  Constant launch speed 

message: I did the derivation of the 'Angle  \theta required to hit coordinate (x,y)'. I belief that there's a v^2 missing in equation (2f) and it should look like that:
: p = \frac{v^2\pm\sqrt{v^4-gv^2(gx^2+2y)}}{gx} 
Can somebody confirm that?
DigDra (talk) 23:59, 1 March 2014 (UTC)
thread_title:  Is Equation 2f correct? 

message: I think the derivations show too many steps, Angle \theta required to hit coordinate (x,y) is quite explicit, same for talking about units for friction.
The Catching section uses a different absurd convention, it use a for 'altitude' and h for velocity. — Preceding unsigned comment added by Timetraveler3.14 (talk • contribs) 22:50, 15 November 2014 (UTC)
thread_title:  Derivations - condensing & notation 

message: Hello fellow Wikipedians,
I have just attempted to maintain the sources on Trajectory of a projectile.  I managed to add archive links to 1 , out of the total 1 I modified, whiling tagging 0 as dead.
Please take a moment to review my changes to verify that the change is accurate and correct.  If it isn't, please modify it accordingly and if necessary tag that source with  to keep Cyberbot from modifying it any further.  Alternatively, you can also add  to keep me off the page's sources altogether.  Let other users know that you have reviewed my edit by leaving a comment on this post.
Below, I have included a list of modifications I've made:
*Added archive http://web.archive.org/web/20071112185623/http://www.dtic.mil:80/doctrine/jel/doddict/data/b/00611.html to http://www.dtic.mil/doctrine/jel/doddict/data/b/00611.html
Cheers.—cyberbot IITalk to my owner:Online 16:10, 5 July 2015 (UTC)
thread_title:  Sources modified on [[Trajectory of a projectile]] 

message: I removed the second part of  as it was completely wrong. In particular, it purported to have a solution for a problem which is known to be unsolved, see "Comments on some recent work by Shouryya Ray" The author makes the incorrect assumption that motion in both axes is independent, which is false for quadratic air resistance since the differential equations of motion are coupled. The first part also needs some work with citations, but it is mathematically accurate. Duxdx (talk) 01:26, 23 November 2015 (UTC)
replies: text: Luckily the French know how to do physics! (:fr:Trajectoire d'un projectile#Trajectoire soumise à la résistance de l'air) --Iste ridiculum vitam est (talk) 14:37, 10 July 2017 (UTC)
thread_title:  Removing Quadratic Air Resistance 

message: In the derivation for the "Trajectory of a projectile with air resistance", high velocity case (air resistance proportional to the square of velocity), there is a mistake in the components of the force, which also leads to wrong conclusions/calculations after that. In the article the force is taken to be F=-k(v_x^2,v_y^2) when it should be F=-k(v_x\sqrt{v_x^2+v_y^2},v_y\sqrt{v_x^2+v_y^2}). Obviously the air resistance should be in opposite direction of the velocity vector (v_x,v_y) which is only the case for the second form of the force. bamse (talk) 11:38, 24 March 2017 (UTC)
replies: text: I don't entirely understand how you arrive at F=-k(v_x\sqrt{v_x^2+v_y^2},v_y\sqrt{v_x^2+v_y^2}) given that \sqrt{v_x^2+v_y^2} = v. Do you have a citation for that derivation? (I'm not disputing that the article might be wrong, but your final formula smells just as wrong, if indeed the current article is wrong.) --Izno (talk) 14:11, 24 March 2017 (UTC)
replies: text: You basically have to satisfy two equations: (1) the magnitude of the force is |\mathbf{F}|=k\mathbf{v}^2 and (2) the direction of the force should be anti-parallel to the direction of the velocity vector, i.e. \mathbf{F}=- c \mathbf{v}, where c is positive (and may depend on v_x, v_y). From the second equation you get that |\mathbf{F}|=c\sqrt{v_x^2+v_y^2}. Equating this expression with the first equation you see that c\sqrt{v_x^2+v_y^2}=k\mathbf{v}^2=k(v_x^2+v_y^2) or c=k\sqrt{v_x^2+v_y^2} which proves that \mathbf{F}=-c(v_x,v_y)=-k(v_x\sqrt{v_x^2+v_y^2},v_y\sqrt{v_x^2+v_y^2}). bamse (talk) 20:49, 24 March 2017 (UTC)

text: I even found a reliable source: Projectile motion with air resistance quadratic in the speed which you can read for free here. bamse (talk) 20:57, 24 March 2017 (UTC)
thread_title:  High velocity air resistance is wrong