Projectile motion

As a simple illustration of the concepts introduced What is the initial vector velocity with which the
in the previous subsections, let us examine the
following problem. Suppose that a projectile is
projectile is launched into the air at (say) ?
As illustrated in Fig. 16, given that the magnitude of

launched upward from ground level, with speed ,

this velocity is , its horizontal component is
making an angle with the horizontal. Neglecting
the effect of air resistance, what is the subsequent directed along the -axis, and its direction
trajectory of the projectile?
Our first task is to set up a suitable Cartesian subtends an angle with this axis, the components
coordinate system. A convenient system is

illustrated in Fig. 16. The -axis points vertically of take the form
upwards (this is a standard convention), whereas
(71
the -axis points along the projectile's initial )
direction of horizontal motion. Furthermore, the
origin of our coordinate system corresponds to the

launch point. Thus, corresponds to ground

level.
Note that has zero component along the -axis,
Neglecting air resistance, the projectile is subject to which points into the paper in Fig. 16.
Since the projectile moves with constant
acceleration, its vector displacement
a constant acceleration , due to
gravity, which is directed vertically downwards.
Thus, the projectile's vector acceleration is written
from its launch point satisfies [see
(70 Eq. (64)]
)
(72
)
Here, the minus sign indicates that the acceleration

is in the minus -direction (i.e., downwards), as

opposed to the plus -direction (i.e., upwards).

Making use of Eqs. (70) and (71), the -, -, and

-components of the above equation are written

(73
)

(74
)

(75
)

Figure 16: Coordinates for the

projectile problem
respectively. Note that the projectile moves with
(78
)
constant velocity, , in the

-direction (i.e., horizontally). This is hardly

surprising, since there is zero component of the
when . In other words, neglecting air
projectile's acceleration along the -axis. Note, resistance, a projectile travels furthest when it is
further, that since there is zero component of the
launched into the air at to the horizontal.

projectile's acceleration along the -axis, and the The maximum altitude of the projectile is
projectile's initial velocity also has zero component

attained when (i.e., when the

along this axis, the projectile never moves in the -
projectile has just stopped rising and is about to
direction. In other words, the projectile's trajectory
start falling). It follows from Eq. (75) that the
is 2-dimensional, lying entirely within the -
plane. Note, finally, that the projectile's vertical
motion is entirely decoupled from its horizontal maximum altitude occurs at time
motion. In other words, the projectile's vertical . Hence,
motion is identical to that of a second projectile
(79
launched vertically upwards, at , with the )

initial velocity (i.e., the initial vertical

velocity component of the first projectile)--both
projectiles will reach the same maximum altitude at Obviously, the largest value of ,
the same time, and will subsequently strike the
ground simultaneously.
(80
Equations (73) and (75) can be rearranged to give )

(76
)
is obtained when the projectile is launched

vertically upwards (i.e., ).

As was first pointed out by Galileo, and is illustrated
in Fig. 17, this is the equation of a parabola. The

horizontal range of the projectile corresponds to

its -coordinate when it strikes the ground (i.e.,

when ). It follows from the above expression

(neglecting the trivial result ) that

(77
)

Figure 17: The parabolic trajectory of a projectile

Note that the range attains its maximum value,
What is a Projectile? acceleration (not a motion). Recall from the Unit 2
that Newton's laws stood in direct opposition to the
In Unit 1 of the Physics Classroom Tutorial, we
common misconception that a force is required to
learned a variety of means to describe the 1-
keep an object in motion. This idea is simply not
dimensional motion of objects. In Unit 2 of the
true! A force is not required to keep an object in
Physics Classroom Tutorial, we learned how
motion. A force is only required to maintain an
Newton's laws help to explain the motion (and
acceleration. And in the case of a projectile that is
specifically, the changes in the state of motion) of
moving upward, there is a downward force and a
objects which are either at rest or moving in 1-
downward acceleration. That is, the object is
dimension. Now in this unit we will apply both
moving upward and slowing down.
kinematic principles and Newton's laws of motion to
understand and explain the motion of objects To further ponder this concept of the downward
moving in two dimensions. The most common force and a downward acceleration for a projectile,
example of an object which is moving in two consider a cannonball shot horizontally from a very
dimensions is a projectile. Thus, Lesson 2 of this high cliff at a high speed. And suppose for a
unit is devoted to understanding the motion of moment that the gravity switch could be turned off
projectiles.A projectile is an object upon which the such that the cannonball would travel in the
only force acting is gravity. There are a variety of absence of gravity? What would the motion of such
examples of projectiles. An object dropped from a cannonball be like? How could its motion be
rest is a projectile (provided that the influence of air described? According to Newton's first law of
resistance is negligible). An object which is thrown motion, such a cannonball would continue in motion
vertically upward is also a projectile (provided that in a straight line at constant speed. If not acted
the influence of air resistance is negligible). And an upon by an unbalanced force, "an object in motion
object is which thrown upward at an angle to the will ...". This is Newton's law of inertia.
horizontal is also a projectile (provided that the Now suppose that the gravity switch is turned on
influence of air resistance is negligible). A projectile and that the cannonball is projected horizontally
is any object which once projected or dropped from the top of the same cliff. What effect will
continues in motion by its own inertia and is gravity have upon the motion of the cannonball?
influenced only by the downward force of gravity. Will gravity affect the cannonball's horizontal
By definition, a projectile has only one force acting motion? Will the cannonball travel a greater (or
upon it - the force of gravity. If there was any other shorter) horizontal distance due to the influence of
force acting upon an object, then that object would gravity? The answer to both of these questions is
not be a projectile. Thus, the free-body diagram of a "No!" Gravity will act downwards upon the
projectile would show a single force acting cannonball to affect its vertical motion. Gravity
downwards and labeled force of gravity (or simply causes a vertical acceleration. The ball will drop
Fgrav). Regardless of whether a projectile is moving vertically below its otherwise straight-line, inertial
downwards, upwards, upwards and rightwards, or path. Gravity is the downward force upon a
downwards and leftwards, the free-body diagram of projectile which influences its vertical motion and
the projectile is still as depicted in the diagram at causes the parabolic trajectory which is
the right. By definition, a projectile is any object characteristic of projectiles.
upon which the only force is gravity.
A projectile is an object upon which the only force is
gravity. Gravity acts to influence the vertical motion
of the projectile, thus causing a vertical
acceleration. The horizontal motion of the projectile
Projectile Motion and Inertia is the result of the tendency of any object in motion
to remain in motion at constant velocity. Due to the
Many students have difficulty with the concept that absence of horizontal forces, a projectile remains in
the only force acting upon an upward moving motion with a constant horizontal velocity.
projectile is gravity. Their conception of motion Horizontal forces are not required to keep a
prompts them to think that if an object is moving projectile moving horizontally. The only force acting
upward, then there must be an upward force. And if upon a projectile is gravity!
an object is moving upward and rightward, there
must be both an upward and rightward force. Their
belief is that forces cause motion; and if there is an Next Section: Characteristics of a Projectile's
upward motion then there must be an upward Trajectory
force. They reason, "How in the world can an object Jump To Lesson 3: Forces in Two Dimensions
be moving upward if the only force acting upon it is
gravity?" Such students do not believe in Projectile
Newtonian physics (or at least do not believe A projectile is any object propelled through space
strongly in Newtonian physics). Newton's laws by the exertion of a force which ceases after
suggest that forces are only required to cause an launch. Although a thrown baseball could be
considered a projectile, the word more often refers explosives are those launched from railguns,
to a weapon. For details of the mathematics coilguns, and mass drivers, as well as kinetic
surrounding projectile trajectory, see equations of energy penetrators. All of these weapons work by
motion. attaining a high muzzle velocity (hypervelocity),
and collide with their objective, converting their
Contents kinetic energy into destructive shock waves and
[hide] heat.
Some kinetic weapons for targeting objects in
• 1 Motive force spaceflight are anti-satellite weapons and anti-
• 2 Non-kinetic effects ballistic missiles. Since they need to attain a high
velocity anyway, they can destroy their target with
• 3 Kinetic projectiles their released kinetic energy alone; explosives are
• 4 Wired projectiles not necessary. Compare the energy of TNT, 4.6
MJ/kg, to the energy of a kinetic kill vehicle with a
• 5 Typical projectile
closing speed of 10 km/s, which is 50 MJ/kg. This
speeds
saves costly weight and there is no detonation to
• 6 Miscellaneous be precisely timed. This method, however, requires
• 7 See also direct contact with the target, which requires a
more accurate trajectory.
• 8 External links
With regard to anti-missile weapons, the Arrow
[edit] Motive force missile and MIM-104 Patriot have explosives, but
the Kinetic Energy Interceptor (KEI), Lightweight
Arrows, darts, spears, and similar weapons are fired Exo-Atmospheric Projectile (LEAP, see RIM-161
using pure mechanical force applied by another Standard Missile 3), and THAAD being developed do
solid object; apart from throwing without tools, not (see Missile Defense Agency).
mechanisms include the catapult, slingshot, and
bow. See also Hypervelocity terminal ballistics,
Exoatmospheric Kill Vehicle (EKV).
Other weapons use the compression or expansion
of gases as their motive force. A kinetic projectile can also be dropped from
aircraft. This is applied by replacing the explosives
Blowguns and pneumatic rifles use compressed of a regular bomb e.g. by concrete, for a precision
gases, while most other guns and firearms utilize hit with less collateral damage. A typical bomb has
expanding gases liberated by sudden chemical a mass of 900 kg and a speed of impact of
reactions. Light gas guns use a combination of 800 km/h (220 m/s). It is also applied for training
these mechanisms. the act of dropping a bomb with explosives. [1] This
Railguns utilize electromagnetic fields to provide a method has been used in Operation Iraqi Freedom
constant acceleration along the entire length of the and the subsequent military operations in Iraq by
device, greatly increasing the muzzle velocity. mating concrete-filled training bombs with JDAM
GPS guidance kits, to attack vehicles and other
Some projectiles provide propulsion during (part of) relatively "soft" targets located too close to civilian
the flight by means of a rocket engine or jet engine.
structures for the use of conventional high
In military terminology, a rocket is unguided, while explosive bombs.
a missile is guided. Note the two meanings of
"rocket": an ICBM is a missile with rocket engines. A kinetic bombardment may involve a projectile
dropped from Earth orbit.
[edit] Non-kinetic effects
A hypothetical kinetic weapon that travels at a
Many projectiles, e.g. shells, contain an explosive significant fraction of the speed of light, usually
charge. With or without explosive charge a
found in science fiction, is termed a relativistic kill
projectile can be designed to cause special vehicle (RKV).
damage, e.g. fire (see also early thermal weapons),
or poisoning (see also arrow poison). [edit] Wired projectiles
[edit] Kinetic projectiles Some projectiles stay connected by a cable to the
launch equipment after launching it:
See also: KE-Munitions
• for guidance: wire-guided missile (range up
Projectiles which do not contain an explosive
to 4000 meters)
charge are termed kinetic projectile, kinetic energy
weapon, kinetic warhead or kinetic penetrator. • to administer an electric shock, as in the
Classic kinetic energy weapons are blunt projectiles case of a Taser (range up to 10.6 meters);
such as rocks and round shot, pointed ones such as two projectiles are shot simultaneously,
arrows, and somewhat pointed ones such as each with a cable.
bullets. Among projectiles which do not contain
 • to make a connection with the target, either pistol) 1224 km/h kJ/kg
to tow it towards the launcher, as with a
whaling harpoon, or to draw the launcher to 12.7x99 mm
the target, as a grappling hook does. (bullet of a
800 m/s, 320
[edit] Typical projectile speeds heavy 2625 ft/s 1790 mph
2880 km/h kJ/kg
machine
See also: Orders of magnitude (speed) and Muzzle gun)
velocity
5.56x45 mm
Kinetic (standard
energy bullet used 920 m/s, 470
Speed 3018 ft/s 2058 mph
density in many 3312 km/h kJ/kg
Projectile (m/s), (ft/s) (mph)
= assault
(km/h)
Speed rifles)
^2 / 2
125x1400 m
4.43 m/s, 1700 m/s, 1.4
object falling m (shell of a 5577 ft/s 3803 mph
15.948 14.5 ft/s 9.9 mph 9.8 J/kg 6120 km/h MJ/kg
1m tank)
km/h
2kg
object falling 14 m/s, Tungsten 3000 m/s,
46 ft/s 31 mph 98 J/kg 4.5
10 m 50.4 km/h Slug (from 10800 9843 ft/s 6711 mph
MJ/kg
Experimental km/h
thrown club Railgun)
(weapon) 40 m/s, 800
130 ft/s 90 mph
(expert 144 km/h J/kg up to
thrower) ICBM reentry up to up to up to 8
13000 ft/
vehicle 4 km/s 9000 mph MJ/kg
s
object falling 45 m/s, 980
150 ft/s 100 mph
100 m 162 km/h J/kg projectile of up to up to up to
up to
a light gas 23000 ft/ 16000 mp 24
refined (= 7 km/s
gun s h MJ/kg
flexible)
45 m/s, 1000
atlatl dart 150 ft/s 100 mph satellite in
162 km/h J/kg 26000 ft/ 19000 mp 32
(expert low earth 8 km/s
thrower) s h MJ/kg
orbit

80-lb-draw closing
58 m/s, Exoatmosph
pistol 1.7 speed ~33000 f ~22000 ~ 50
208.8 190 ft/s 130 mph eric Kill
crossbow kJ/kg roughly t/s mph MJ/kg
km/h Vehicle
bolt 10 km/s

paintball 91 m/s, projectile

4.1
fired from 327.6 300 ft/s 204 mph (e.g. space
kJ/kg closing
marker km/h debris) and ~53000 f ~36000 ~ 130
speed 0 -
target both t/s mph MJ/kg
175-lb-draw 97 m/s, 16 km/s
4.7 in low earth
crossbow 349.2 320 ft/s 217 mph orbit
kJ/kg
bolt km/h

air gun pellet 100 m/s, [edit] Miscellaneous

328 ft/s 224 mph 5 kJ/kg
6 mm BB 360 km/h Ballistics analyze the projectile trajectory, the
forces acting upon the projectile, and the impact
rifle bullet 150 m/s, 11 that a projectile has on a target. A guided missile is
492 ft/s 336 mph
4.5 mm 540 km/h kJ/kg not called a projectile.
An explosion, whether or not by a weapon, causes
air gun pellet 244 m/s,
29.8 the debris to act as multiple high velocity
(conventiona 878.4 800 ft/s 545 mph
kJ/kg projectiles. An explosive weapon, or device may
l maximum) km/h
also be designed to produce many high velocity
projectiles by the break-up of its casing, these are
9x19 mm 340 m/s, 1116 ft/s 761 mph 58
correctly termed fragments.
(bullet of a
Projectile is also the name of an annual anarchist
• 4 External links
film festival based in Newcastle UK * [2]
Trajectory [edit] Physics of trajectories
A trajectory is the path a moving object follows A familiar example of a trajectory is the path of a
through space. The object might be a projectile or a projectile such as a thrown ball or rock. In a greatly
satellite, for example. It thus includes the meaning simplified model the object moves only under the
of orbit - the path of a planet, an asteroid or a influence of a uniform homogenous gravitational
comet as it travels around a central mass. A force field. This can be a good approximation for a
trajectory can be described mathematically either rock that is thrown for short distances for example,
by the geometry of the path, or as the position of at the surface of the moon. In this simple
the object over time. approximation the trajectory takes the shape of a
parabola. Generally, when determining trajectories
In control theory a trajectory is a time-ordered set
it may be necessary to account for nonuniform
of states of a dynamical system (see e.g. Poincaré
gravitational forces, air resistance (drag and
map). In discrete mathematics, a trajectory is a
aerodynamics). This is the focus of the discipline of
ballistics.
sequence of values calculated by the One of the remarkable achievements of Newtonian
iterated application of a mapping f to an element x mechanics was the derivation of the laws of Kepler,
of its source. in the case of the gravitational field of a single point
mass (representing the Sun). The trajectory is a
conic section, like an ellipse or a parabola. This
agrees with the observed orbits of planets and
comets, to a reasonably good approximation.
Although if a comet passes close to the Sun, then it
is also influenced by other forces, such as the solar
wind and radiation pressure, which modify the orbit,
and cause the comet to eject material into space.
Newton's theory later developed into the branch of
theoretical physics known as classical mechanics. It
employs the mathematics of differential calculus
(which was, in fact, also initiated by Newton, in his
youth). Over the centuries, countless scientists
contributed to the development of these two
disciplines. Classical mechanics became a most
prominent demonstration of the power of rational
Illustration showing the trajectory of a bullet fired at thought, i.e. reason, in science as well as
an uphill target. technology. It helps to understand and predict an
enormous range of phenomena. Trajectories are but
one example.
Contents
Consider a particle of mass m, moving in a potential
[hide] field V. Physically speaking, mass represents
inertia, and the field V represents external forces,
• 1 Physics of trajectories
of a particular kind known as "conservative". That
• 2 Examples is, given V at every relevant position, there is a way
○ 2.1 Uniform gravity, no drag or wind to infer the associated force that would act at that
position, say from gravity. Not all forces can be
 2.1.1 Derivation of the expressed in this way, however.
equation of motion
The motion of the particle is described by the
 2.1.2 Range and height second-order differential equation
 2.1.3 Angle of elevation
○ 2.2 Uphill/downhill in uniform gravity
in a vacuum
with
 2.2.1 Derivation based on
equations of a parabola
○ 2.3 Orbiting objects
• 3 See also
On the right-hand side, the force is given in terms Now translating back to the inertial frame the co-
ordinates of the projectile becomes y = xtan(θ) −
of , the gradient of the potential, taken at g(x / vh)2 / 2 That is:
positions along the trajectory. This is the
mathematical form of Newton's second law of
motion: mass times acceleration equals force, for
such situations.
,
[edit] Examples
(where v0 is the initial speed, h is the height the
[edit] Uniform gravity, no drag or wind projectile is fired from, and g is the acceleration due
to gravity).

Trajectories of three objects thrown at the same [edit] Range and height
angle (70°). The black object doesn't experience The range, R, is the greatest distance the object
any form of drag and moves along a parabola. The travels along the x-axis in the I sector. The initial
blue object experiences Stokes' drag, and the green velocity, vi, is the speed at which said object is
object Newton drag. launched from the point of origin. The initial
angle, θi, is the angle at which said object is
The ideal case of motion of a projectile in a uniform released. The g is the respective gravitational pull
gravitational field, in the absence of other on the object within a null-medium.
forces(such as air drag), was first investigated by
Galileo Galilei. To neglect the action of the
atmosphere, in shaping a trajectory, would have
been considered a futile hypothesis by practical
minded investigators, all through the Middle Ages in
Europe. Nevertheless, by anticipating the existence
The height, h, is the greatest parabolic height said
of the vacuum, later to be demonstrated on Earth
object reaches within its trajectory
by his collaborator Evangelista Torricelli[citation needed],
Galileo was able to initiate the future science of
mechanics.[citation needed] And in a near vacuum, as it
turns out for instance on the Moon, his simplified
parabolic trajectory proves essentially correct.
In the analysis that follows we derive the equation [edit] Angle of elevation
of motion of a projectile as measured from an
inertial frame, at rest with respect to the ground, to In terms of angle of elevation θ and initial speed v:
which frame is associated a right-hand co-ordinate
system - the origin of which coincides with the point
of launch of the projectile. The x-axis is parallel to
the ground and the y axis perpendicular to it giving the range as
( parallel to the gravitational field lines ). Let g be
the acceleration of gravity. Relative to the flat
terrain, let the initial horizontal speed be vh =
vcos(θ) and the initial vertical speed be vv =
vsin(θ). It will also be shown that, the range is 2vhvv This equation can be rearranged to find the angle
for a required range
/ g, and the maximum altitude is ; The
maximum range, for a given initial speed v, is
obtained when vh = vv, i.e. the initial angle is 45
degrees. This range is v2 / g, and the maximum (Equation II: angle of
altitude at the maximum range is a quarter of that. projectile launch)

[edit] Derivation of the equation of motion Note that the sine function is such that there are
two solutions for θ for a given range dh. Physically,
Assume the motion of the projective is being this corresponds to a direct shot versus a mortar
measured from a Free fall frame which happens to shot up and over obstacles to the target. The angle
be at (x,y)=(0,0) at t=0. The equation of motion of θ giving the maximum range can be found by
the projectile in this frame ( by the principle of considering the derivative or R with respect to θ
equivalence) would be y = xtan(θ). The co- and setting it to zero.
ordinates of this free-fall frame, with respect to our
inertial frame would be y = − gt2 / 2. That is, y = −
g(x / vh)2 / 2.
which has a non trivial solutions at
Equation 11 may also be used to develop the
. The maximum range is then "rifleman's rule" for small values of α and θ (i.e.
close to horizontal firing, which is the case for many
. At this angle sin(π / 2) = 1 so the firearm situations). For small values, both tanα and
tanθ have a small value and thus when multiplied
together (as in equation 11), the result is almost
zero. Thus equation 11 may be approximated as:
maximum height obtained is .
To find the angle giving the maximum height for a
given speed calculate the derivative of the
maximum height H = v2sin(θ) / (2g) with respect to
And solving for level terrain range, R

θ, that is which is zero "Rifleman's rule"

when . So the maximum height Thus if the shooter attempts to hit the level
distance R, s/he will actually hit the slant target. "In
other words, pretend that the inclined target is at a
horizontal distance equal to the slant range
distance multiplied by the cosine of the inclination
is obtained when the projectile is fired
angle, and aim as if the target were really at that
straight up.
horizontal position."[1]
[edit] Uphill/downhill in uniform gravity in a
[edit] Derivation based on equations of a
vacuum
parabola
Given a hill angle α and launch angle θ as before, it
The intersect of the projectile trajectory with a hill
can be shown that the range along the hill Rs forms
may most easily be derived using the trajectory in
a ratio with the original range R along the
parabolic form in Cartesian coordinates (Equation
imaginary horizontal, such that:
10) intersecting the hill of slope m in standard
linear form at coordinates (x,y):

(Equation (Equation 12) where in this

11) case, y = dv, x = dh and b = 0

In this equation, downhill occurs when α is between Substituting the value of dv = mdh into Equation 10:
0 and -90 degrees. For this range of α we know:
tan( − α) = − tanα and sec( − α) = secα. Thus for
this range of α, Rs / R = (1 + tanθtanα)secα. Thus
Rs / R is a positive value meaning the range
downhill is always further than along level terrain.
The lower level of terrain causes the projectile to
remain in the air longer, allowing it to travel further
horizontally before hitting the ground.
While the same equation applies to projectiles fired (Solving
uphill, the interpretation is more complex as above x)
sometimes the uphill range may be shorter or
longer than the equivalent range along level This value of x may be substituted back into the
terrain. Equation 11 may be set to Rs / R = 1 (i.e. linear equation 12 to get the corresponding y
the slant range is equal to the level terrain range) coordinate at the intercept:
and solving for the "critical angle" θcr:
 and provided much of the motivation for the
development of differential calculus.
Range of a projectile

Now the slant range Rs is the distance of the

intercept from the origin, which is just the
hypotenuse of x and y:

The path of this projectile launched from a height y0

has a range d.

In physics, a projectile launched with specific initial

conditions in a uniform gravity field will have a
Now α is defined as the angle of the hill, so by predictable range. As in Trajectory of a projectile,
definition of tangent, m = tanα. This can be we will use:
substituted into the equation for Rs: • g: the gravitational acceleration—usually
taken to be 9.80 m/s2 near the Earth's
surface
• θ: the angle at which the projectile is
launched

Now this can be refactored and the trigonometric • v: the velocity at which the projectile is
launched
• y0: the initial height of the projectile
identity for may be used:
• d: the total horizontal distance travelled by
the projectile
When neglecting air resistance, the range of a
projectile will be

Now the flat range R = v2sin2θ / g = 2v2sinθcosθ / g

by the previously used trigonometric identity and
sinθ / cosθ = tanθ so:
If (y0) is taken to be zero, meaning the object is
being launched on flat ground, the range of the
projectile will then simplify to

[edit] Orbiting objects

If instead of a uniform downwards gravitational Contents
force we consider two bodies orbiting with the [hide]
mutual gravitation between them, we obtain
Kepler's laws of planetary motion. The derivation of • 1 Derivations
these was one of the major works of Isaac Newton ○ 1.1 Flat Ground
 Note that when (θ) is 45°, the solution becomes
○ 1.2 Uneven Ground
○ 1.3 Maximum Range on Uneven
Ground

[edit] Derivations
[edit] Uneven Ground
[edit] Flat Ground
Now we will allow (y0) to be nonzero. Our equations
First we examine the case where (y0) is zero. The of motion are now
horizontal position (x(t)) of the projectile is

and
In the vertical direction

Once again we solve for (t) in the case where the

We are interested in the time when the projectile (y) position of the projectile is at zero (since this is
returns to the same height it originated at, thus how we defined our starting height to begin with)

By applying the quadratic formula Again by applying the quadratic formula we find
two solutions for the time. After several steps of
algebraic manipulation

or

The square root must be a positive number, and

The first solution corresponds to when the projectile since the velocity and the cosine of the launch
is first launched. The second solution is the useful angle can also be assumed to be positive, the
one for determining the range of the projectile. solution with the greater time will occur when the
Plugging this value for (t) into the horizontal positive of the plus or minus sign is used. Thus, the
equation yields solution is

Applying the trigonometric identity Solving for the range once again

sin(x + y) = sin(x)cos(y) + sin(y)cos(x)

If x and y are same,

sin(2x) = 2sin(x)cos(x)
[edit] Maximum Range on Uneven Ground
allows us to simplify the solution to
It might be of interest to know how to compute the
elevation angle which will provide the maximum
range when launching the projectile from a non-
zero initial height. This can be computed by finding
the derivative of the range with respect to the
elevation angle and setting the derivative to zero to
find the extremum:

where and R = horizontal range.

Setting the derivative to zero provides the

equation:

Substituting u = (cosθ)2 and 1 − u = (sinθ)2

produces:

Which reduces to the surprisingly simple

expression:

Replacing our substitutions yields the angle that

produces the maximum range for uneven ground,
ignoring air resistance:

Note that for zero initial height, the elevation angle

that produces maximum range is 45 degrees, as
expected. For positive initial heights, the elevation
angle is below 45 degrees, and for negative initial
heights (bounded below by y0 > − 0.5v2 / g), the
elevation angle is greater than 45 degrees.
Example: For the values g = 9.80m / s2, y0 = 40m ,
and v = 50m / s, an elevation angle θ = 41.1015°
produces a maximum range of Rmax = 292.11