Trajectory of a projectile with linear drag

Summary

Dans cette vidéo, l'analyse du trajectoire d'un projectile soumis à une force de trainée proportionnelle à sa propre vitesse est présentée. Après avoir établi les équations paramétriques de x et y en fonction du temps, la vidéo combine ces équations pour obtenir une équation cartésienne de la trajectoire. La trajectoire, affectée par la résistance de l'air, perd sa symétrie typique pour une trajectoire sans air resistance. Les conditions initiales et la résolution des équations différentielles permettent de trouver les expressions de x(t) et y(t), et finalement, une équation de trajectoire en fonction de x révèle une forme moins symétrique et plus verticale à mesure que le projectile avance.

Takeaways

• 🔍 L'analyse porte sur la trajectoire d'un projectile soumis à une force de drag proportionnelle à sa propre vitesse.
• 📐 La trajectoire est décrite par des équations paramétriques en x et y en fonction du temps, qui sont ensuite combinées pour obtenir une équation cartésienne.
• 📉 La présence de la résistance de l'air altère la forme de la trajectoire, qui n'est plus symétrique comme dans le cas sans résistance.
• ⏳ L'application de la deuxième loi de Newton en forme vectorielle permet d'établir un lien entre les forces agissant sur le projectile et son accélération.
• 🧮 Les équations différentielles pour x et y sont résolues en utilisant des méthodes standard, notamment en cherchant des solutions exponentielles pour l'équation sans force de gravitation.
• 🔄 La solution générale pour x en fonction du temps est une combinaison linéaire de deux solutions, l'une constante et l'autre décroissant exponentiellement.
• 🎯 Les conditions initiales (position et vitesse initiales) sont appliquées pour déterminer les constantes dans les équations de la trajectoire.
• 📉 L'équation différentielle pour y inclut une composante de force de gravitation, qui nécessite une solution en deux parties : la fonction complémentaire et l'intégrale particulière.
• 📌 La forme de la trajectoire est affectée par la résistance de l'air, qui ralentit le composant horizontal de la vitesse, rendant la trajectoire plus vertical à mesure que le projectile avance.
• 🧐 L'équation cartésienne finale de la trajectoire montre une dépendance en logaritme de la distance x, ce qui reflète la diminution de la vitesse horizontale due à la résistance de l'air.

Q & A

Quelle est la force de trainée (drag force) décrite dans le script ?

La force de trainée est une force proportionnelle à la vitesse de l'objet, notée comme -b * v, où b est une constante et v est la vitesse du projectile.

Quels sont les deux principaux composants de la force qui agit sur le projectile dans le script ?

Les deux principaux composants de la force sont la gravité, qui agit垂直向下, et la force de trainée (air resistance force), qui est proportionnelle à la vitesse du projectile.

Comment le script utilise-t-il la deuxième loi de Newton pour analyser le mouvement du projectile ?

Le script utilise la deuxième loi de Newton, qui relie la force à l'accélération, pour établir des équations différentielles pour les composantes x et y de la position du projectile.

Quelle est la forme vectorielle de l'équation de la deuxième loi de Newton appliquée au projectile ?

La forme vectorielle est F = m * a, où F représente la force resultante, m est la masse du projectile, et a est l'accélération, notée comme r'', où r est le vecteur de position.

Comment le script décompose l'équation vectorielle en composantes x et y ?

Le script décompose l'équation en deux équations séparées pour les composantes x et y, en prenant en compte la force de trainée et la gravité, et en utilisant les notations x'', y'', x', et y' pour les dérivées temporelles des coordonnées.

Quelle est la solution générale pour la composante x du mouvement du projectile ?

La solution générale pour x(t) est une combinaison linéaire de deux solutions: une constante (a) et une fonction exponentielle décroissante (b * e^(-b*t/m)), où a et b sont des constantes déterminées par les conditions initiales.

Comment le script résout l'équation différentielle pour la composante y en prenant en compte la force de trainée et la gravité ?

Le script résout l'équation pour y(t) en trouvant d'abord la fonction complémentaire (solution homogénéisée), puis en ajoutant une intégrale particulière basée sur une conjecture informée pour tenir compte de la force de trainée et de la gravité.

Quels sont les deux principaux termes dans la solution générale pour y(t) ?

Les deux principaux termes sont la fonction complémentaire, qui est une fonction exponentielle, et l'intégrale particulière, qui est une combinaison linéaire de t et de t², ajustée pour équilibrer la force de trainée et la gravité.

Comment le script utilise-t-il les conditions initiales pour déterminer les constantes dans les équations de mouvement ?

Le script applique les conditions initiales (position et vitesse au début du mouvement) pour résoudre les équations et trouver les valeurs des constantes a et b dans les expressions x(t) et y(t).

Quelle est la forme finale de l'équation cartésienne pour le trajectoire du projectile avec la force de trainée ?

L'équation cartésienne finale est y = (b * u_y + mg / b) * x - (m^2 * g / b^2) * ln(1 - b * x / m * u_x), où b est la constante de trainée, u_x et u_y sont les composantes initiales de la vitesse, et g est l'accélération due à la gravité.

Outlines

00:00

🚀 Introduction au trajectoire d'un projectile avec résistance de l'air

Le paragraphe introduit la notion d'étude de la trajectoire d'un projectile soumis à une force de drag ou de résistance de l'air proportionnelle à sa propre vitesse. L'objectif est de trouver des équations paramétriques pour x et y en fonction du temps, puis de les combiner pour obtenir une équation cartésienne de la trajectoire. La présence de la résistance de l'air altère la forme de la trajectoire par rapport au cas sans résistance. L'analyse débute par l'application de la deuxième loi de Newton pour établir un lien entre les forces agissantes sur le projectile, y compris la gravité et la force de drag, et son mouvement.

05:00

🔍 Application de la deuxième loi de Newton et équations différentielles

Ce paragraphe se concentre sur l'application de la deuxième loi de Newton sous forme vectorielle au projectile, ce qui conduit à une équation différentielle pour décrire son mouvement. L'accent est mis sur la résolution de cette équation pour la composante x, en prenant en compte l'absence de force de résistance dans cette direction. Une solution de substitution est proposée, basée sur une fonction exponentielle, pour laquelle les conditions initiales du mouvement sont appliquées pour déterminer les constantes dans l'équation.

10:01

🧮 Calcul des constantes et résolution de l'équation pour la composante x

Le paragraphe présente la méthode pour calculer les constantes dans l'équation de la composante x du mouvement du projectile. En utilisant les conditions initiales, notamment la vitesse initiale et la position de départ, les valeurs de ces constantes sont déterminées. La solution complète pour x en fonction du temps est ensuite obtenue en substituant ces valeurs dans l'équation antérieurement trouvée, ce qui permet d'exprimer la position horizontale du projectile en fonction du temps.

15:04

📉 Résistance de l'air et équation différentielle pour la composante y

Le paragraphe traite de la résolution de l'équation différentielle pour la composante y du mouvement du projectile, qui inclut la force de gravité et la force de résistance de l'air. La solution est divisée en deux parties : la fonction complémentaire et l'intégrale particulière. Après avoir trouvé la forme générale de la fonction complémentaire, une forme de l'intégrale particulière est proposée en fonction de la force de gravité constante. Les constantes de cette solution sont calculées en substituant les conditions initiales pour la vitesse et la position verticales.

20:07

🌐 Équation cartésienne de la trajectoire avec résistance de l'air

Le paragraphe conclut l'analyse en éliminant la variable temporelle pour exprimer la trajectoire en tant qu'équation cartésienne de y en fonction de x. Cela est réalisé en utilisant la solution pour la composante x pour calculer le temps, puis en le substituant dans l'équation pour la composante y. Le résultat est une équation qui décrit la trajectoire verticale en fonction de la position horizontale, montrant l'effet de la résistance de l'air qui rend la trajectoire moins symétrique et plus inclinée vers le bas à mesure que le projectile avance.

25:07

📉 Forme de la trajectoire affectée par la résistance de l'air

Le dernier paragraphe discute de la forme de la trajectoire du projectile, soulignant comment la présence de la résistance de l'air la rend moins symétrique et plus verticale à mesure que le projectile avance. L'importance de la composante horizontale de la vitesse qui diminue progressivement est mise en évidence, ce qui explique la pente croissante de la trajectoire vers le bas à la fin du mouvement du projectile.

Keywords

💡Trajectoire

La trajectoire fait référence au chemin que suit un objet, tel qu'un projectile, dans l'espace et le temps. Dans la vidéo, l'analyse de la trajectoire est le fil conducteur principal, car elle illustre comment le projectile se déplace sous l'influence des forces telles que la gravité et la résistance de l'air. L'exemple clé est la trajectoire asymétrique du projectile en raison de la force de trainée, qui contraste avec la trajectoire plus classique et symétrique sans résistance de l'air.

💡Force de trainée

La force de trainée est une résistance qui s'oppose au mouvement d'un objet dans un fluide, comme l'air. Elle est décrite dans le script comme une force proportionnelle à la vitesse du projectile, notée -b * v. Cette force est cruciale pour comprendre la trajectoire modifiée du projectile, car elle affecte sa vitesse horizontale et verticale tout au long de son vol.

💡Paramètres

Les paramètres sont des valeurs qui décrivent la configuration initiale d'un système, comme la position et la vitesse d'un projectile au moment de son lancement. Dans le script, les paramètres tels que la position initiale et la vitesse initiale du projectile sont utilisés pour établir les conditions initiales pour les équations de mouvement.

💡Équations paramétriques

Les équations paramétriques sont des équations qui expriment les coordonnées d'un point en fonction d'une variable supplémentaire, souvent le temps. Dans la vidéo, les équations paramétriques x(t) et y(t) sont utilisées pour décrire la position du projectile en fonction du temps, ce qui est essentiel pour comprendre son mouvement dans le plan XY.

💡Équation cartésienne

Une équation cartésienne est une relation entre les coordonnées x et y d'un point dans un plan sans référence au temps. Dans le script, l'objectif est de combiner les équations paramétriques pour obtenir une équation cartésienne, qui décrit la forme de la trajectoire du projectile dans l'espace sans la variable temporelle.

💡Newton's Second Law

La deuxième loi de Newton relie les forces appliquées à un objet à son accélération, exprimée par la formule F=ma. Dans le script, cette loi est utilisée pour établir les équations de mouvement du projectile, en tenant compte de la force de trainée et de la gravité.

💡Système de coordonnées

Un système de coordonnées est un cadre de référence pour décrire la position des objets dans l'espace. Dans le script, le système de coordonnées est défini avec l'origine au point de lancement du projectile, avec les axes x et y représentant les directions horizontale et verticale, respectivement.

💡Vitesse

La vitesse est une mesure de la quantité de changement de position d'un objet par unité de temps. Dans le script, la vitesse du projectile est décomposée en composantes horizontale et verticale, ce qui est essentiel pour résoudre les équations de mouvement dans les deux directions.

💡Force gravitationnelle

La force gravitationnelle est la force attractive exercée par la Terre sur les objets, qui s'exprime dans le script par le terme -mg, où m est la masse du projectile et g est l'accélération due à la gravité. Cette force est constante et垂直e, influençant la composante verticale de l'accélération du projectile.

💡Équations différentielles ordinaires

Les équations différentielles ordinaires sont des équations qui impliquent des dérivées d'une seule variable, généralement le temps. Dans le script, les équations de mouvement du projectile sont des équations différentielles ordinaires qui sont résolues pour trouver les fonctions de position x(t) et y(t).

Highlights

Analyzing the trajectory of a projectile under the influence of air resistance.

Drag force is proportional to the velocity of the projectile.

Parametric equations x(t) and y(t) will be derived to describe the projectile's path.

The trajectory shape is asymmetrical due to air resistance, contrasting with no air resistance cases.

Newton's second law is applied to relate forces to the projectile's motion.

Forces considered include gravity and air resistance, modeled as -b*v.

Vector notation is used to express the forces acting on the projectile.

Component form of the vector equation simplifies the analysis into x and y components.

Two separate differential equations for x and y motion are established.

Solving the x-component equation leads to an exponential function of time.

Initial conditions are applied to determine constants in the x(t) equation.

The y-component equation includes a forcing term due to gravity.

The solution for y(t) includes both a complementary function and a particular integral.

Constants for the y(t) equation are determined using initial velocity and position.

The final parametric equations for x(t) and y(t) are derived.

Elimination of the time variable t results in a Cartesian equation for the trajectory.

The derived Cartesian equation shows the impact of air resistance on the projectile's path.

The trajectory's shape is influenced by the decreasing horizontal velocity due to drag.

Transcripts

hello everyone in this video we're going

to look at the trajectory of a

projectile which is experiencing a drag

force or an air resistance force

proportional to its own velocity in

other words it's going to be filling a

drag force f subscript track which is

minus some constant which i'm calling b

times the velocity v of that projectile

okay now in my last video i did a very

similar thing in the case of no air

resistance which turns out to be

mathematically much easier

but just as a reminder of what we're

going to be doing

we'll first find parametric equations x

and y as a function of time

and then when we've done that we're

going to combine them together to get

a cartesian equation for the trajectory

of this projectile and you can already

see from the diagram that i've drawn

it's it no longer has this nice

symmetrical shape

that the trajectory of a particle has

when there's no air resistance okay

and again just to set up the problem

i'm going to say that the coordinate

system has its origin at the starting

point of the projectile the y-coordinate

increases in the vertical direction and

the x coordinate increases

as you move to the right and the the

initial velocity

of this projectile is going to be a

vector which i'm going to call u

okay so there's the setup now let's

think about how we're going to actually

analyze this so

because we've got air resistance we're

going to have to think about forces

and if we're trying to link forces to

well how an object moves we want to be

using newton's second law right because

newton's second law gives us a

connection between force and

acceleration okay so let's think about

what happens if we apply newton's second

law in vector form to this projectile

right so

the second noise says force equals mass

times acceleration well the acceleration

if we call the mass m um then

the the mass times acceleration time is

just m times

r double dot where r is the position

vector

so it's a standard notation for position

vector and a double dot means two time

derivatives so this r double dot is just

the acceleration all right so the other

side of this equation needs to be the

resultant force now there are two forces

acting on this projectile there's

gravity acting straight straight down

right um

the size of that gravitational force is

going to be

mg so the mass times the gravitational

field strength it's going to have a

minus sign because it's pointing

downwards

and i'm going to say that it's in the

y-hat direction in other words the unit

vector in the y-direction okay

now there is also the the drag or the

air resistance force

which is going to be basically this term

that i wrote up there minus b v

but using this kind of position uh

notation instead of v i'm going to write

r dot because the velocity is the first

time derivative of the position right so

we're going to have a minus b

and then r dot there okay

so

this is a vector equation which makes it

a bit tricky um to get a clearer idea of

what's what this actually means i'm

going to write it out in component form

so

the r double dot vector is basically x

double dot and y double dot right um and

then on the right hand side we've got

minus mg well the unit vector in the y

direction is just 0 1

and this r dot vector um is

x dot and then y dot okay so i've just

kind of written all of the vectors out

in terms of their components

and this actually gives us two equations

right one from the x components and one

from the y components right so what we

get is the following two equations

the

x equation is just m x double dot is

equal to minus b x dot right because

there's just a zero in the x component

of that

first term on the right hand side and

the y equation is going to be m y double

dot is equal to

minus mg

minus b y dot like that okay

now i'm going to label these one and two

and if we want to know what the

trajectory looks like we basically have

to solve those two equations to find x

as a function of time and y as a

function of time okay so let's look at

those equations one by one let's start

with the x direction so we're going to

look at equation one

now if we rearrange this into a more

standard form for uh for an ordinary

differential equation we want to get a

zero on one side right and so if i put

all the terms on the same side

and divide through by the mass we can

put that first equation into the form x

double dot plus b over m x dot is equal

to zero so this is like a kind of

standard form for for a second order

ordinary differential equation

um and we can also use standard methods

for this um

which

essentially involves guessing a solution

um

the trial solution that we're going to

try

uh is going to be as follows so we want

to try a solution which is proportional

to e to the lambda t the reason for that

is because we're looking for a function

such that when you differentiate it

twice and add it to a multiple of the

first derivative you get zero and the

only types of functions that behave that

way are

exponentials

or signs and causes but you can write

exponentials and signs and causes in

terms of each other so

um you can you can choose either one

uh i'm going for for exponentials in

this case okay and this lambda is just

some constant that we're going to have

to solve for so if we substitute this

equation e to the lambda 2 sorry this

this tri solution e to the lambda t into

the differential equation

uh well what happens to the x double dot

is that

you pull down a factor of lambda squared

right because each time you

differentiate e to the lambda t with

respect to t you pull down a factor of

lambda and so

you get a factor of lambda squared

um and then you've still got this e to

the lambda t

the

second term this b over m x dot there's

just going to be a single factor of

lambda like that so that's going to be

lambda um

e to the lambda t but we've got to make

sure that's been multiplied by b over m

right and that is supposed to equal 0.

now

because um

e to the lambda t itself can never be

zero we can basically just divide

through by that

and then we've got a quadratic equation

for lambda that we have to solve um

if we factorize the resulting quadratic

we can take out a lambda and get lambda

brackets lambda plus

b over m

is equal to zero okay now

this has two solutions

one of them is lambda is 0 from this

pre-factor and one of them is lambda is

minus b over m from the bracketed term

and so because there are two solutions

to that equation

there are two valid solutions for x as a

function of time and the most general

solution will be a linear combination of

those two solutions all right based on

that

we can say our x is going to be some

constant a

times well basically e to the 0 because

one of the solutions was lambda

equals 0 but e to the 0 is just 1 and so

this just becomes a constant a

then from our other solution for lambda

we get some other constant b

times e to the minus b

t over m right because this corresponds

to lambda being minus b over m um

but then the solution itself was e to

the lambda t right so we get e to the

minus b t over m

now um if we differentiate that to find

um the

velocity or the the x component of the

velocity x dot um the a term would

disappear you would bring down a factor

of minus b over m

from the second term and so this is

minus small b over m times capital b

times e to the minus b t of rem okay

now the reason i did this the reason i

differentiated that to find the x

component of the velocity is that that

allows us to apply some initial

conditions right we have two initial

conditions we know that

the projectile is starting from the

origin so in other words x is zero when

t is zero but we also know that the

initial velocity is this u vector and so

we have an initial condition on the

speed in both the x and y directions as

well okay so

if we apply those boundary conditions

let's think about the velocity one first

so x dot when time when the time is zero

is just the x component of the u vector

right so x dot of zero is u subscript x

now if we use this expression for x dot

um

and substitute in t is zero and x dot is

equal to ux what do we get well we find

u x is just minus small b over m times

capital b and then this exponential term

becomes one because it becomes e to the

zero when t is 0 right

so we get this equation which we can

just rearrange and solve for capital b

so we find that that b

is equal to

minus m

use

u subscript x divided by small b like

that okay so we found one of the

constants

to find the constant a we can use our

other boundary condition which is that x

when t is zero is equal to zero right so

if we do that substitute that into the

first equation x as a function of time

you get 0 equals a

plus

b and then you would have an e to the 0

here but again e to the 0 is 1 so a plus

b is 0 which means a has to be minus b

in other words a has to just be

m

u subscript x over small b like that so

same thing but just it's positive

instead of negative okay

um

and so we found the constants capital a

and capital b

all we need to do to find x as a

function of time then is substitute

those back in to our expression for x as

a function of time

if we do that we get x is equal to where

we've got this m

u subscript x over b

this is going to be a common factor in

both the a term and the b term right

because remember a and b are the same

just give or take a minus sign so it can

actually take out as a factor this mu x

over b

and then i'm going to have 1 minus e to

the minus b t over m in brackets there

okay so that's just come from

substituting a and b into

that expression that we derived earlier

by solving the differential equation so

there we go we now know what x is as a

function of time

so let's go through a similar procedure

and solve for y as a function of time

okay

so our differential equation

um

that tells us how y behaves with

equation two um

if we rearrange that equation into a

standard form

well the left hand side will be the same

pretty much as the x equation you would

get y double dot plus b over m

um

y dot

um but then on the right hand side we

would have an extra term left over which

is just minus g okay um so this we can

kind of think of this as like a forcing

term

um

and

because of this minus g there is

something else we've got to think about

when we're solving this differential

equation

so again i'm not entering this video to

be a detailed uh

introduction to how to solve ordinary

differential equations so i'll go

through this fairly quickly basically

the solution consists of two parts the

complementary function and the

particular integral now the

complementary function

is basically what you would get if you

solved this equation but with a zero on

the right hand side instead of a minus g

okay so

because of the fact that

the left-hand side of this equation has

exactly the same form as the x equation

we can immediately say that the

complementary function which i'm going

to call y subscript cf has to be of the

same form

a um plus

b e to the minus b t over m so these a

and b's

the the this a and this b are not

necessarily the same as this a and this

b um

but there you go it's going to have the

same form right

but the difference is we've got to add

on

another contribution which is this thing

called the particular integral now the

way to find the particular integral is

to basically make

make a guess or an informed guess based

on the form of what you've got on the

right hand side of your equation

and um and also well what you've got on

the left side of your equation as well

now

because the right hand side of the

equation in other words the forcing term

is just a constant

and because on the left hand side i have

a first derivative and a second

derivative i'm going to

guess a a form for the particular

integral as

basically being a second order

polynomial right so i'm going to call it

alpha t squared

plus

beta t right because if you

differentiate um

a second order or quadratic polynomial

twice

you get a constant which is the same

form as what we've got on the right hand

side there right so that's where that's

coming from i'm not bothering to put a

constant term gamma there because

um

well you don't you don't have y

appearing directly on the left hand side

of this differential equation right so

if you differentiate this expression y

particular integral the gamma would just

disappear and so

we know that it's it's just going to be

zero in the end anyway okay so i'm just

trying this general form

of the the particular integral um so all

we've got to do having made that

informed guess as to the functional form

of the particular integral is take that

and substitute it back in now

if we

differentiate this twice

the

first term

will become

just 2 alpha right because if we

differentiate it once we get 2 out for t

differentiate again you get 2 alpha the

second term would disappear

right because if you differentiate once

you get beta and then again and it

disappears because it's a constant and

so

the y double dot for the particular

integral would just be

2 alpha okay then for the y dot term

well we get b plus b over m um

and then we've got that two alpha t from

differentiating the first term once and

then we also get plus b over m uh times

beta which comes from differentiating

the second term once okay and all of

that stuff is supposed to be equal to

minus g

so

then what we want to do is just compare

coefficients

of the different types of term on the

left hand side and the right hand side

we can immediately see that alpha

actually has to be zero because

the there is a t term here b over m

times two alpha t but there is no t term

on the right hand side right and so

the only way to

eliminate the t dependence on the left

hand side is for alpha to actually be

zero right so we can immediately say

that alpha is supposed to be zero

um

and then we can compare the constant

terms right we um we find that well the

constant term on the left hand side is 2

alpha plus b over m times beta the

constant term on the right hand side is

minus g we already know that the alpha

term is actually zero and so

we're just equating this b over m times

beta with this minus g now if those are

going to be equal what that means is

that beta

is supposed to be

minus mg over b and there we go we found

that alpha is zero beta is minus mg over

b

okay

so

putting all of that together um

our overall solution for y

um is going to be well the complementary

function plus the particular integral

okay complementary function was a plus

b e to the minus b t over m

um then we've got to add the particular

integral

if we substitute these values of alpha

and beta that we just found back into

that equation for the particular

integral alpha t squared plus b to t

we are just left with

minus

mg t over b right from that from that b

to term there okay so this is the

general solution for for y um and again

so that we can apply boundary conditions

i'm also going to differentiate that

once to find y dot right so if we do

that

as we did with the

the x um

earlier on

back up there

the

the first part of this will just

differentiate to give us well minus b

over m times capital b

e to the minus b t over m then this

particular integral part

will just turn into a constant because

well that's a linear term and so we get

a minus mg over b there

so let's apply our boundary conditions

and find uh what the values of this a

and this b are okay so we know that the

initial component of velocity in the y

direction in other words uh y dot

at t equals zero is just the y component

of the u vector in other words u

subscript y okay so if we take our

expression for y dot substitute t equals

zero in

and see what happens well we get um

u y

uh is supposed to be equal to

uh minus b over m times capital b

minus mg over b again because e to the

zero is one and so that term just

becomes one okay now if we rearrange

this um to make b the subject we find

that b is supposed to be

minus m over b and then we're going to

have a bracketed term which will be uy

plus

mg over b right so just rearrange the

previous line to make capital b the

subject so we know b

let's then apply our other boundary

condition which is that it starts at the

origin again so y the y coordinate at t

equals zero is simply zero

if we substitute that in um we find that

so we're substituting that into this

equation for y as a function of time

if y is 0 we just get 0 equals then

there is going to be an a nothing

changes there the b term capital b term

is just going to be b because e to the 0

is 1. this linear term disappears

because we're substituting t equals 0.

so again a plus b is 0. so a is just the

same as b but with a positive sign

instead of a negative sign so a

is going to be

m over small b times u y

plus mg over b so now we've found our

constants capital a and capital b

um putting all of that together

we can then get y as a function of time

y is going to be

again there's this common factor of m

over b

in brackets u y plus

mg over b

um

and then um well we get a

we've kind of factored out this term

if you look at the form of this

expression up here

for y as a function of time

the first term is just a which is

exactly that so if i factored that out

we need to have a one in brackets here

and then from the b term we get minus e

to the minus b t

over m okay so we've got that bit and

then

we've got our linear part at the end

minus

t

over b

which didn't depend on the constants

capital a and capital b so there we go

now we've got a parametric equation for

x as a function of time up at the top

and a parametric equation for y as a

function of time at the bottom there so

the last thing we want to do is try to

eliminate t uh the time variable and

just express y as a function of x

because that will tell us the um the

shape or the cartesian equation

of the trajectory okay now

the way i'm going to do that is that

notice that t only appears once in

equation one up at the top here right so

we can actually

fairly easily rearrange that top

equation involving x to make t the

subject right so let's do that take

equation one

um multiply by b over mu x so the left

hand side is going to become b x over m

u subscript x

then we're just left with this one minus

e to the

minus b t over m on the right hand side

okay

if we do one minus both sides

we get e to the minus b t

over m

is just one minus

b x over m

u x okay

and so

we would have to take logs of both sides

right to to get t on its own so if we

take logs of both sides

and then multiply up by this factor

that's in front of t right which is the

the minus b over m

we are going to get that t is minus um

m over b

all right we've got to flip that factor

upside down because we're basically

dividing by b over m uh then we get the

natural log

of the entire right hand side so minus m

over b

ln of 1 minus bx over m u subscript x

okay and there we go now we've expressed

the time t

in terms of the x coordinate

and so

if we're trying to eliminate that time

all we've got to do is take this

expression for t and substitute it into

equation 2 at the top there for y right

so let me just write that down so we're

basically substituting that

into equation 2

and what's going to happen well we get

y equals we've still got this first bit

um m over b

u y

plus

m g over b

um

now

we have this bracket to term 1 minus e

to the minus bt over m but we know that

that's the same from uh

this

intermediate step here we know that that

entire bracketed term 1 minus e to the

minus b 2 over m is the same as bx over

mu subscript x so here i can just put a

b x over

m

u subscript x like that okay

now

we've still got to deal with that linear

term at the end we get minus mg over b

and then we just substitute directly

this expression for t

and so we just put here minus m over b

and then

ln of 1 minus

bx

over

mu subscript x

okay so we are very nearly there we've

just got to simplify this as much as we

can there's no time dependence left in

here which is good um

so

first thing i'm going to point out is

that we've got an m over b here and a b

over m there so those are going to

cancel so i'm going to cross those out

um

and so what are we left with well we've

got this u y

plus

mg

over b

then that whole thing is going to be

over u x

and then we've still got that x on the

top of the fraction there so there's

going to be an x there

now for the second term we've got a

minus and a minus so it's going to end

up being positive we've got

m squared

g

over b squared

and then we've got this same uh log term

right so

ln of

1 minus

bx over mu x

and

i guess

personally i would want to remove this

fraction within a fraction we can do

that just by multiplying up by b so our

final cartesian equation for the

trajectory of a projectile with um

linear drag

is going to be as follows so y equals

b

times u subscript y plus

mg divided by

b

u subscript x because we had to times

everything in that first fraction by b

to remove the fraction within a fraction

that is all multiplied by x

and then well nothing's changed about

that second term so i'm just going to

copy and paste that and put it there

okay

um

so notice it doesn't have this nice

symmetrical shape that the trajectory

had when there was no air resistance

basically this log term

at the end here makes the trajectory

kind of start falling off

or kind of moving downwards quite

quickly as it reaches the end of its

trajectory

which makes sense because basically

what's happened is that the

um

horizontal component of the uh the

velocity is getting smaller and smaller

um because of because of the the air

resistance right so the trajectory kind

of gets more and more close to being

vertical as the as the projectile

reaches the end of its motion