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Projectile Motion
Definition:
body of mass m launched with speed v0 at angle
from the horizontal;
air resistance FPres
= bvP , b = nonnegative
constant (possibly zero)
x
z
F = mg
m
zv0
Prerequisites:
fundamentals of Newtonian mechanics
motion in one dimension, with and without res-
istance
Why study it?
a very simple dynamical system with an exactsolution in closed
form;
occurs frequently in everyday applications
Summary:
The bodys position as a function of time is
x(t)=
m
b v0cos
1 exp
b
m t
z(t)= mg
bt+
m
b
v0sin +
mg
b
1 exp
b
mt
Go to derivation.
Go to MAPLE code
Go to Java applet
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We are now going to analyze a standard problemin elementary
physics: projectile motion. Sup-pose we launch a projectile with a
fixed speed v0
at some angle from the earths surface. Whatpath does it follow,
and at what angle must welaunch it, in order for it to travel the
maximumdistance along the earths surface?
For simplicity, well first solve the problem inthe artificial
case in which there is no air
resistance, and then well go on to the moregeneral case.
Although we have not yet studied thegravitational interaction,
you may be familiarwith the relevant fact: near earths surface,
all
bodies (whatever their inertial mass) have an
acceleration of approximately 9.81 m s2 in thedownward
direction. This is true as long as theeffects of air resistance are
negligible. This valueis given a special symbol, g, and is called
thegravitational acceleration at earths surface:
g 9.81 m s2.
Let's orient our coordinates so that the z-axis
points upwards. Then the force due to gravity is
FP= mgz ,
wherez is the unit vector in the z direction. The
minus sign indicates that the force is directeddownwards. This
force is constant - does not
depend on time, position, or velocity.
x
z
F = mg
m
zv
0
v0
cos
v0
sin
The force of gravity is always in the negative z-
direction - it has no component in thex-direction.
Accordingly, the problem splits into two separateones, both of
which we have already solved whenwe considered one-dimensional
motion.
z-direction: motion with constant acceleration
mz = mg .
x-direction: motion with constant velocity
mx = 0 .
(Recall that the double dot notation stands forsecond time
derivative).
We choose coordinates so that the initial valuesofx andz are
both zero. The initial velocities are
the components of the velocity vector shown inthe last
figure:
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z(0) = v0 sin and x (0) = v0 cos .
The solutions to the equations of motion withthese initial
conditions are
x(t) = v0 cos t and
z(t) = v0 sin t1
2g t2 .
If we are interested only in the shape of the
trajectory, we eliminate tfrom these equations:
z =x tan g
2v20cos2
x2 .
This is a parabola opening downwards: aparabolic arch. (This
expression forz is valid
as long as /2 - when the projectile is firedvertically. In that
case,x remains zero.)
How far does the projectile go, before hitting theground? To
answer this, we must find thevalue(s) ofx for which z=0. Of course,
one
solution is x=0 - the point of launch, but we are
not interested in that one. The other one is easilyfound to
be
x() =v20 sin 2
g.
From this, we find that the maximum distance
xmax
=v20
gis reached when =
4.
Lets plot the paths for several different values of (all with
the same initial speed):
x
z
xmax
We see that the trajectory that starts out at =/4- the red one -
goes the farthest. Trajectories thatstart out at smaller dont go as
high or as far,and trajectories that start at larger go higher,but
still not as far.
The resources for this section contain a movie
comparing various trajectories. There is also amovieillustrating
the fact that the velocity in the
horizontal direction is constant, while that in thevertical
direction is not.
Exercise: firing projectile up a hill
Suppose you are firing a projectile up a hill. Atwhat angle to
the horizontal should you fire it,
http://www.scar.utoronto.ca/~pat/fun/NEWT3D/MOVIES/PRJCTILE.MOVhttp://www.scar.utoronto.ca/~pat/fun/NEWT3D/MOVIES/PROJCOMP.MOVhttp://www.scar.utoronto.ca/~pat/fun/NEWT3D/MOVIES/PROJCOMP.MOVhttp://www.scar.utoronto.ca/~pat/fun/NEWT3D/MOVIES/PROJCOMP.MOVhttp://www.scar.utoronto.ca/~pat/fun/NEWT3D/MOVIES/PRJCTILE.MOV
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so that it goes the maximum distance along thesurface of the
hill?
D()
The answer is
= 4
+ 2
.
(Hint: show that
D() =2v20 cos sin
g cos2.
Note that this expression goes over into ourprevious expression
forx() when =0.)
Example: shooting the monkey
Heres a classic example that every beginningphysics student has
traditionally learned. A
hunter sees a monkey in a tree, and decides toshoot it. He knows
that this particular species of
monkey always falls from the tree at the instant
the shot is fired. At what angle must the hunter
aim, in order to hit the monkey?
x
z
1
1
1
1
1
1
1
0
0
0
0
0
0
0
1 1 1 1 1 1 1 1 1 1 1 10 0 0 0 0 0 0 0 0 0 0 0
monkey
D
H
The first step is to solve for the time at whichimpact occurs.
Thex-component of the velocity
of the projectile is v0
cos , and the distance in
thex-direction isD, so the time of impact is
t=
D
v0 cos .
At the time of impact, the z-coordinates of the
monkey and bullet must be the same:
H1
2
g t2 = v0 sin t1
2
g t2 .
The terms involving the acceleration cancel out
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(both bodies accelerate at the same rate -Galileos experiment!),
and therefore
t=H
v0 sin .
Equating the two expressions for the time ofimpact yields
tan =H
D.
This says that the hunter should aim directly at
the monkey! The downward acceleration of the
monkey exactly compensates for the downwardacceleration of the
bullet, as long as they startfalling at the same time.
It may at first seem a bit surprising that this resultdoesnt
depend on the speed of the bullet, but itmakes sense when you think
about it. The fasterthe bullet, the less time the bullet has to
fall, butthe monkey also has less time to fall, so the speedhas no
effect.
The resources for this section contain a movieillustrating the
manner in which the monkeymeets his maker. (Disclaimer: no actual
monkeydied to make this movie.)
Inclusion of air resistance
In many physical cases, there is some resistanceto motion. For
example, a body could be sliding
on a surface with friction present. Or, a bodycould be moving
near the earth, with airresistance.
In many cases, the force resisting the motion isproportional to
the velocity of the body.
Mathematically, this is written as
FPres
= bvP,
where b is a positive constant. The minus sign
indicates that the force resists the motion, so isdirected
opposite to the velocity.
We would like to illustrate the procedure ofsolving Newtons law
when such a force isinvolved. Just to make it interesting, we
will
suppose that the total force on the body is thesum of a constant
force FP and the above resistive
force. This will make our analysis immediatelyapplicable to the
case of projectile motion in auniform gravitational field.
It is always a good idea to use physical intuition
to get an idea of the nature of the solution, beforebeginning
the mathematics. In the present case, itmakes it easy to see one
important aspect of thesolution without doing a lot of
calculations.
As time goes on, the external constant force will just balance
the resistive force, giving zero net
force. The body will then move with a constantvelocity called
the terminal velocityv
t.
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b
v
P
t FP
vP t
The above figure shows the equal and oppositeforces in red. The
net force is
FPnet
=FP bvPt= 0 ,
which gives for the terminal velocity
vPt=
FP
b .
Lets solve the equation of motion and see howthis is reflected
in the solution. Newton's lawreads
m
dvP(t)
dt =FP
bvP
(t),
which we have written entirely in terms of thevelocity and its
first derivative.
The solution is analogous to the one-dimensionalcase. The result
is
vP(t) =FP
b+
vP0
FP
b
exp
b
mt
,
where vP0 is the velocity at t=0. As t becomes
large, the second term vanishes and the velocity
approaches FP /b, as we know it should. Noticethat the second
term never actually becomes zeroat any finite time - it just gets
closer and closer.
Integrating once again, we find that the positionvector is given
by
xP(t) =xP0 + FP
bt+ m
b
vP0 F
P
b
1 exp
b
mt
,
wherexP0 is the position vector at t=0.
Application: projectile motion with air resistance
Lets go back and examine projectile motion, thistime including
air resistance. With ourcoordinates oriented in the same way as
before,
the constant force due to gravity is FP= mgz , andwe find that
the above vector equation gives twoseparate equations:
x(t)=m
bv0cos
1 exp
b
mt
z(t)= mg
bt+
m
b
v0sin +
mg
b
1 exp
b
mt
.
Here is some MAPLEcode which will generate
the above solutions:
eq1:=(D@@2)(x)(t)=-(b/m)*D(x)(t):eq2:=(D@@2)(z)(t)=-g-(b/m)*D(z)(t):dsolve({eq1,x(0)=0,D(x)(0)=v[0]*cos(theta)},x(t));dsolve({eq2,z(0)=0,D(z)(0)=v[0]*sin(theta)},z(t));
If all we are interested in is the shape of the
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trajectory, we should eliminate the time andexpressz directly in
terms ofx. Solving for the
time from the x-equation, which we can do as
long as /2, we get
t= m
bln
1
bx
mv0cos
.
Inserting this into thez-equation, we obtain
z =
m2g
b2 ln
1
bx
mv0cos
+
sin +
mg
bv0
x
cos .
At this point, we begin to realize how tedious it isto keep
writing all the constants. It is moreefficient to express
everything in dimensionlessform. To do this, we have to decide
on
appropriate units.For our unit of length, it is sensible to take
themaximum horizontal distance the projectile cantravel with no air
resistance; from before, weknow this is v20 /g. Dividing by this
unit, we get
dimensionless variables
=g
v02x and =
g
v02z .
For a dimensionless quantity that indicates theamount of damping
present, it is sensible to takethe ratio of the initial speed to
the terminal speed.
From before, we know that the latter is mg/b, soour
dimensionless damping coefficient is
/v0
vt
=bv0
mg.
This quantity is zero when no damping is present.In terms of
these dimensionless variables, theequation for the shape of the
trajectory reads
=1
2ln
1
c
+
s+
1
c ,
where c and s are short for cos and sin,respectively. This form
of the equation is mucheasier to work with than the original.
It turns out that the maximum horizontal distanceis no longer
attained when the launch angle is 45
degrees, but rather something less than that. Inthe next plot,
the red trajectory has a 45-degreelaunch angle and the blue one,
launched at asmaller angle with the same speed, goes farther:
1/2
1
no resistance
with resistance
(=1)
(=0)
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Also shown for comparison is the 45-degreetrajectory with no air
resistance. We see how thetrajectory with resistance becomes skewed
by thedecreasing horizontal velocity. The trajectorywith no
resistance is symmetric (a parabolicarch). We also see that the
horizontal distancecovered is less when resistance is present. This
iswhat you might expect.
Lets find out how much less the distance is. To
find how far the projectile goes at a given angle, we set =0 and
solve for . One solution is=0, of course, but we are not interested
in thatone. The other solution is more complicated, andcant be
expressed in closed form. In general, itmust be obtained
numerically.
However, we can obtain an approximate solutionin the case where
the resistance is small. Then,we can expand in powers of . (For
practice inexpansions like this, see the section on tech-niques for
checking answers.) We will keepcorrection terms linear in . That
means that we
must keep terms up to the cubic term in theexpansion of the
logarithm:
0 .1
2
c
1
2
c
2
1
3
c
3
+
s+
1
c.
The terms of order 1/ cancel out, and we may
multiply the rest by c/, obtaining
0 = s
2c
2
3c2.
We know from before that the answer when =0is
x =2v20sincos
g, or = 2sc .
This agrees with the equation we just obtained.Lets write the
answer when the damping is
nonzero as
. 2sc (1 + )
and solve for the coefficient . We find
= 4
3
s .
Therefore, we have the result for the horizontaldistance
travelled at a given angle :
. sin 2
1
4
3sin
What is the maximum horizontal distancepossible, and for what
angle is this attained? Tofind the maximum, we set the derivative
to zero:
0 . 2 cos 2
1
4
3
sin
4
3
sin2cos.
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The answer when =0 is = /4. So we write
.
4+
and solve for the coefficient . First, we workout
cos
2+ 2
= sin 2 .2 .
Inserting this and dropping all but linear terms:
0 . 4 4
3 cos
4.
Hence,
=
1
3 2 .
So the maximum distance is attained when thelaunch angle is
.
4
3 2 .
Note that this is slightly less than /4, the valuewhen the air
resistance is zero. This may or maynot be surprising, depending on
how good yourphysical intuition is.
To find out the maximum distance, we insert this
back into the expression for . First, we work out
sin
2+ 2
= cos 2 . 1 .
Hence, the maximum distance is
1H
1
4
3 sin
4
,
and we have a simple result for the maximum
horizontal distance attainable:
max
.1 4
3 2 .
Of course, this is less than 1, the value with no
resistance.
