Talk:Trajectory of a projectile

The free body diagram
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)


 * 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)

its good too B madan11 (talk) 16:24, 22 February 2016 (UTC)

This article doesn't belong in Wikipedia
It's a high school problem... solved... (No trolling intended... just being clear) --euyyn 22:54, 25 July 2006 (UTC)

I believe the air resistance calculation is incorrect:
is not directly proportional to the velocity of the object, it is directly proportional to the square of the velocity of the object. Imagine the object moving through air at v, the object is colliding with n air molecule per unit of time t. Each collision will exert an average amount of impulse per molecule i. The force of drag on the object will be ni/t=drag. If the speed is increased to 2v, the molecules strike the object twice as hard, imparting approximately 2 time the impulse 2i per impact. At the same time, twice the number of molecules 2n are hitting the object in the same amount of time t. (2i)(2n/s)=four times the drag. Unfortunately, this means that someone who is better than me at WikiMath formatting will have to redo the differential equations. 69.76.242.48 06:56, 13 October 2006 (UTC)


 * The matter must have to be with fluids. We should maybe consider viscosity to understand it (the air just around the projectile moves along the surface, not just colliding and "bouncing"). --euyyn 04:46, 19 December 2006 (UTC)


 * Drag is proportional to v at low velocities and v^2 at high velocities. --Gellender 06:21, 20 December 2006 (UTC)


 * Yes, in fact it is only proportional to v for very slow bodies, very small bodies, or for a big viscosity (which isn't the case with air). For further info see Reynolds number and Viscous flow. But I wont fix it since I believe this article doesn't belong in Wikipedia. Maybe in Wikiversity; undergraduate first course of Physics or first course of Differential Equations. --euyyn 21:00, 1 February 2007 (UTC)


 * The example in the end is not well-suited. For the given data drag due to viscosity (proportional to v) should be around 0.001 N and due to pressure around 22 N (proportional to v^2). I guess both regimes could be considered, however surely more concisely. Wrwrwr 20:26, 31 August 2007 (UTC)


 * For a typical projectile (e.g. bullet or cannon projectile) the drag (as said above) is proportional to v squared, and not v. This makes the resulting equation only solvable by numerical means, hence the early military use of computers to calculate trajectories. In the simple case without rotation, no wind, constant air pressure, temperature and gravity, you get two coupled differential equations for velocity in x and y directions. Assuming drag is proportional to V^2, we get:


 * dVx/dt = -k*V^2*cos(flight angle) = -k*V^2*Vx/sqrt(Vx^2+Vy^2) (Pythagoras)
 * dVy/dt = -mg -k*V^2*sin(flight angle) = -mg -k*V^2*Vy/sqrt(Vx^2+Vy^2)
 * or (since V^2 = Vx^2+Vy^2),
 * dVx/dt = -k*Vx*sqrt(Vx^2+Vy^2)
 * dVy/dt = -mg -k*Vy*sqrt(Vx^2+Vy^2)


 * To my knowledge there is no simple elementary solution, but it is easy to numerically calculate on a computer. At any rate, the current article is misleading and should be either deleted or make clear that this is for fluids or at very low speeds and sizes.


 * For the simple case with a projectile fired straight up, the equation is easily solvable. Solutions yield Vgoingup(t)~tan(kt) and Vfallingdown(t)~tanh(kt). —Preceding unsigned comment added by 71.141.139.19 (talk) 08:25, 13 August 2009 (UTC)

A formula for drawing the trajectory on Cartesian coordinates
$$m=\frac{h+d}{d^2}+(\theta\,-1)\frac{1}{d}$$

$$f(x) = -mx^2+\theta\,x+h$$ 

h = launch height

d = flight distance

θ = launch angle (slope)

m = coefficient

You'll get a pretty trajectory (without air resistance in mind, of course) of your favorite projectile on a graphical calculator. Is this awesome, y/n? Well, at least this kind of math is fun. -- nlitement [talk] 01:16, 23 September 2007 (UTC)


 * What is given above is an approximation of the correct Cartesian (parabolic) trajectory equation. From elementary two-dimensional kinematics we can write, under the assumption of constant accelerations a,


 * $${{x}_{t}}\,\,=\,\,{{x}_{0}}+\,\,\,\left[ {{v}_{0}}\cos \left( {{\theta }_{0}} \right) \right]\,\,t\,\,+\,\,\frac{1}{2}{{a}_{x}}\,{{t}^{2}}\,\,\,\,\,\,\,\,\,\,{{y}_{t}}\,=\,\,{{y}_{0}}+\,\,\left[ {{v}_{0}}\sin \left( {{\theta }_{0}} \right) \right]t\,\,\,+\,\,\,\frac{1}{2}{{a}_{y}}\,{{t}^{2}}$$


 * where the subscript zero indicates the initial (i.e., t = 0) values of the displacements x and y (y is positive upward), the (time-varying) velocity magnitude v and the angle of its (time-varying) direction θ; g is the local gravitational acceleration. The angle is measured from the local horizontal, positive counterclockwise. Usually the coordinate system is defined such that the initial x displacement is zero. There is no acceleration in the x (horizontal) direction, and the acceleration in the y (vertical) direction is –g (negative because it is directed downward). With these assumptions we then have the projectile motion in a vacuum parametric equations


 * $${{x}_{t}}\,\,=\,\,\,\left[ {{v}_{0}}\cos \left( {{\theta }_{0}} \right) \right]\,\,t\,\,\,\,\,\,\,\,\,\,\,\,{{y}_{t}}\,\,=\,\,{{y}_{0}}+\,\,\left[ {{v}_{0}}\sin \left( {{\theta }_{0}} \right) \right]t\,\,\,-\,\,\,\frac{1}{2}g\,{{t}^{2}}$$


 * To obtain the rectangular or Cartesian coordinate relation y(x), we eliminate the time parameter t in these equations. Using the xt equation since it is simpler, we see at once that


 * $$t\,\,=\,\,\frac{{{v}_{0}}\cos \left( {{\theta }_{0}} \right)}$$


 * and placing this in the yt equation yields


 * $$y(x)\,\,=\,\,\,\left\{ \frac{-g}{2{{\left[ {{v}_{0}}\cos \left( {{\theta }_{0}} \right) \right]}^{2}}} \right\}{{x}^{2}}\,\,\,+\,\,\,\tan \left( {{\theta }_{0}}\right)\,x\,\,\,+\,\,{{y}_{0}}$$


 * which is the desired expression. This can readily be entered into a graphing calculator.


 * The approximation above depends on knowing the range R, but this is very rarely given in problem statements. Also, R seems to be treated as independent of the initial angle θ0 and the initial height y0. But it can be shown that


 * $$R=\,\,\frac{g}\cos \left( {{\theta }_{0}} \right)\left\{ \,{{v}_{0}}\sin \left( {{\theta }_{0}} \right)\,\,\,+\,\,\,\sqrt{\,{{\left[ {{v}_{0}}\sin \left( {{\theta }_{0}} \right) \right]}^{2}}\,\,+\,\,2\,g\,{{y}_{0}}}\, \right\}$$


 * which uses the usually-given variables initial height, initial velocity magnitude, initial velocity angle. One would, apparently, need to use this range equation to find R before using the above approximation, and it is obvious that R is not independent of these several variables. In other words, we cannot pick an arbitrary R and also an equally arbitrary angle θ to use in the m equation; the range and the angle are related, not independent.


 * We can see in the approximation from the use of the angle θ, which, although it is not stated, must be in radians, that some approximation for the trig functions must have been used. These approximations are generally only reasonable for “small” angles. Indeed, if the approximation given above is plotted along with the correct y(x) equation, the differences are clear. At larger angles (more than, say, 30 degrees or so), the approximation is not very good.


 * In summary, there is no advantage to using this approximation. Simply plot the y(x) function given above in these comments using your calculator; quite a lot of useful analysis can be done directly there (e.g., the range, the maximum height, the velocity at various points). Note that many calculators, especially the TI 83/84, can plot parametric equations, so that the trajectory could also be plotted using the second pair of kinematic equations above. However, in parametric mode some curve analysis options are not available, so using the y(x) form is probably better.


 * Most of this material appears in readily-available undergraduate (and high school) physics textbooks, although it seems that no such references are cited in any of the Wiki articles I’ve seen on this subject matter. There are several projectile-related articles, that, IMHO, should be brought together in a more organized manner, with better exposition, references, and examples (graphics).


 * If you have questions or comments about this material, please put them on my talk page... Regards, Rb88guy (talk) 19:09, 28 September 2009 (UTC)

Tagged with regarding equations
I tagged the section with  instead of cleanup. 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)

Negative exponent?
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)


 * You won't. Meters per second, per second acceleration is indeed ms-2. (User) Wolfkeeper (Talk) 20:59, 19 February 2009 (UTC)


 * 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)

Remove lengthy explanation
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)

Negative gravity?
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)


 * 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)

Things to add to the article
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)

Ballistic spacecraft re-entry
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)

Question about height/range at 45 degrees
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)


 * 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)

These formulas clearly break down as the angle approaches 90 degrees (up)
--184.21.215.174 (talk) 18:23, 14 March 2013 (UTC)


 * How so? DOwenWilliams (talk) 00:58, 15 March 2013 (UTC)


 * I'm sorry, break down may have been the wrong word. I can't recall. But basically they are not useful for calculating anything when the trajectory is vertical (the equations basically collapse to a point, including division by zero, in that case.) So there should be an extra section, or caveats scattered here and there. Or just a disclaimer that defines a "projectile" as being non-vertical and recommends a better article for that. I know Wikipedia is not a textbook, but it is very tempting to quickly gleam equations from. I was able to piece together what I needed from three separate articles including this one. --184.21.215.174 (talk) 14:38, 18 March 2013 (UTC)


 * "Approaches" is the wrong word. The equations fail only when the angle is exactly 90 degrees. But yes. Some mention of this might be useful. DOwenWilliams (talk) 15:11, 18 March 2013 (UTC)


 * In a perfect platonic world that is true. But anytime machines are doing computations precision becomes a real issue. So approaches is correct in that sense.


 * Also you must handle the vertical case as a completely separate case with these equations because you can't pull any meaningful information from them in that case. So any complete solution would have to be conditional on the angle and the equations here would be of no use to figure out the odd case. Other equations on other pages here do not suffer from this. --184.21.215.174 (talk) 15:56, 19 March 2013 (UTC)


 * Equation to get initial velocity on 90 degree angle, given height, should be $$v_{0} = \sqrt{2gh}$$ 201.231.132.215 (talk) 07:40, 31 October 2014 (UTC)

Animated GIF in section Angle required to hit coordinate (x,y)
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)

Constant launch speed
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)

Is Equation 2f correct?
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)

Derivations - condensing & notation
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)

Sources modified on Trajectory of a projectile
Hello fellow Wikipedians,

I have just attempted to maintain the sources on Trajectory of a projectile. I managed to add archive links to 1 1 source, 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 II  Talk to my owner :Online 16:10, 5 July 2015 (UTC)

Removing Quadratic Air Resistance
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)
 * Luckily the French know how to do physics! (fr:Trajectoire d'un projectile) --Iste ridiculum vitam est (talk) 14:37, 10 July 2017 (UTC)

Idea for another section - projectile motion onto an inclined plane/slope
If projectile is fired on an incline/slope of beta (angle between -pi/2 and pi/2) then range is (v^2/g) (sec^2 beta sin(2 theta - beta) - tan beta sec beta). Perhaps this could be included in the article (with overview of derivation).

High velocity air resistance is wrong
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)
 * 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)
 * 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)
 * 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)