Trajectory of a projectile with linear drag
Summary
TLDRDans 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
🚀 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.
🔍 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.
🧮 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.
📉 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.
🌐 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.
📉 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.
Mindmap
Keywords
💡Trajectoire
💡Force de trainée
💡Paramètres
💡Équations paramétriques
💡Équation cartésienne
💡Newton's Second Law
💡Système de coordonnées
💡Vitesse
💡Force gravitationnelle
💡Équations différentielles ordinaires
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
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