# Projectile motion

(Redirected from Ballistic trajectory)
Parabolic water motion trajectory
Components of initial velocity of parabolic throwing

Projectile motion is a form of motion experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the action of gravity only (in particular, the effects of air resistance are passive and assumed to be negligible). This curved path was shown by Galileo to be a parabola, but may also be a line in the special case when it is thrown directly upwards. The study of such motions is called ballistics, and such a trajectory is a ballistic trajectory. The only force of mathematical significance that is actively exerted on the object is gravity, which acts downward, thus imparting to the object a downward acceleration towards the Earth’s center of mass. Because of the object's inertia, no external force is needed to maintain the horizontal velocity component of the object's motion. Taking other forces into account, such as aerodynamic drag or internal propulsion (such as in a rocket), requires additional analysis. A ballistic missile is a missile only guided during the relatively brief initial powered phase of flight, and whose remaining course is governed by the laws of classical mechanics.

Ballistics (Greek: βάλλειν, romanizedba'llein, lit.'to throw') is the science of dynamics that deals with the flight, behavior and effects of projectiles, especially bullets, unguided bombs, rockets, or the like; the science or art of designing and accelerating projectiles so as to achieve a desired performance.

Trajectories of a projectile with air drag and varying initial velocities

The elementary equations of ballistics neglect nearly every factor except for initial velocity and an assumed constant gravitational acceleration. Practical solutions of a ballistics problem often require considerations of air resistance, cross winds, target motion, varying acceleration due to gravity, and in such problems as launching a rocket from one point on the Earth to another, the rotation of the Earth. Detailed mathematical solutions of practical problems typically do not have closed-form solutions, and therefore require numerical methods to address.

## Kinematic quantities of projectile motion

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. This is the principle of compound motion established by Galileo in 1638,[1] and used by him to prove the parabolic form of projectile motion.[2]

The horizontal and vertical components of a projectile's velocity are independent of each other.

A ballistic trajectory is a parabola with homogeneous acceleration, such as in a space ship with constant acceleration in absence of other forces. On Earth the acceleration changes magnitude with altitude and direction with latitude/longitude. This causes an elliptic trajectory, which is very close to a parabola on a small scale. However, if an object was thrown and the Earth was suddenly replaced with a black hole of equal mass, it would become obvious that the ballistic trajectory is part of an elliptic orbit around that black hole, and not a parabola that extends to infinity. At higher speeds the trajectory can also be circular, parabolic or hyperbolic (unless distorted by other objects like the Moon or the Sun). In this article a homogeneous acceleration is assumed.

### Acceleration

Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is constant, being equal to ${\displaystyle \mathbf {v} _{0}\cos \theta }$. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to g.[note 1] The components of the acceleration are:

${\displaystyle a_{x}=0}$,
${\displaystyle a_{y}=-g}$.

### Velocity

Let the projectile be launched with an initial velocity ${\displaystyle \mathbf {v} (0)\equiv \mathbf {v} _{0}}$, which can be expressed as the sum of horizontal and vertical components as follows:

${\displaystyle \mathbf {v} _{0}=v_{0x}\mathbf {\hat {x}} +v_{0y}\mathbf {\hat {y}} }$.

The components ${\displaystyle v_{0x}}$ and ${\displaystyle v_{0y}}$ can be found if the initial launch angle, ${\displaystyle \theta }$, is known:

${\displaystyle v_{0x}=v_{0}\cos(\theta )}$,
${\displaystyle v_{0y}=v_{0}\sin(\theta )}$

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity changes linearly,[note 2] because the acceleration due to gravity is constant. The accelerations in the x and y directions can be integrated to solve for the components of velocity at any time t, as follows:

${\displaystyle v_{x}=v_{0}\cos(\theta )}$,
${\displaystyle v_{y}=v_{0}\sin(\theta )-gt}$.

The magnitude of the velocity (under the Pythagorean theorem, also known as the triangle law):

${\displaystyle v={\sqrt {v_{x}^{2}+v_{y}^{2}}}}$.

### Displacement

Displacement and coordinates of parabolic throwing

At any time ${\displaystyle t}$, the projectile's horizontal and vertical displacement are:

${\displaystyle x=v_{0}t\cos(\theta )}$,
${\displaystyle y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$.

The magnitude of the displacement is:

${\displaystyle \Delta r={\sqrt {x^{2}+y^{2}}}}$.

Consider the equations,

${\displaystyle x=v_{0}t\cos(\theta ),y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$.

If t is eliminated between these two equations the following equation is obtained:

${\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}$.

Since g, θ, and v0 are constants, the above equation is of the form

${\displaystyle y=ax+bx^{2}}$,

in which a and b are constants. This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical.

If the projectile's position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v0 in the aforementioned parabolic equation:

${\displaystyle v_{0}={\sqrt {{x^{2}g} \over {x\sin 2\theta -2y\cos ^{2}\theta }}}}$.

### Displacement in polar coordinates

The parabolic trajectory of a projectile can also be expressed in polar coordinates instead of Cartesian coordinates. In this case, the position has the general formula

${\displaystyle r(\phi )={\frac {v_{0}^{2}\cos ^{2}\theta }{g}}\left(-\tan \phi \sec \phi +{\sqrt {(\tan ^{2}\phi +\tan ^{2}\theta )\sec ^{2}\phi }}\right)}$.

In this equation, the origin is the midpoint of the horizontal range of the projectile, and if the ground is flat, the parabolic arc is plotted in the range ${\displaystyle 0\leq \phi \leq \pi }$. This expression can be obtained by transforming the Cartesian equation as stated above by ${\displaystyle y=r\sin \phi }$ and ${\displaystyle x=r\cos \phi }$.

## Properties of the trajectory

### Time of flight or total time of the whole journey

The total time t for which the projectile remains in the air is called the time of flight.

${\displaystyle y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$

After the flight, the projectile returns to the horizontal axis (x-axis), so ${\displaystyle y=0}$.

${\displaystyle 0=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$
${\displaystyle v_{0}t\sin(\theta )={\frac {1}{2}}gt^{2}}$
${\displaystyle v_{0}\sin(\theta )={\frac {1}{2}}gt}$
${\displaystyle t={\frac {2v_{0}\sin(\theta )}{g}}}$

Note that we have neglected air resistance on the projectile.

If the starting point is at height y0 with respect to the point of impact, the time of flight is:

${\displaystyle t={\frac {d}{v\cos \theta }}={\frac {v\sin \theta +{\sqrt {(v\sin \theta )^{2}+2gy_{0}}}}{g}}}$

As above, this expression can be reduced to

${\displaystyle t={\frac {v\sin {\theta }+{\sqrt {(v\sin {\theta })^{2}}}}{g}}={\frac {v\sin {\theta }+v\sin {\theta }}{g}}={\frac {2v\sin {\theta }}{g}}={\frac {2v\sin {(45)}}{g}}={\frac {2v{\frac {\sqrt {2}}{2}}}{g}}={\frac {{\sqrt {2}}v}{g}}}$

if θ is 45° and y0 is 0.

### Maximum height of projectile

Maximum height of projectile

The greatest height that the object will reach is known as the peak of the object's motion. The increase in height will last until ${\displaystyle v_{y}=0}$, that is,

${\displaystyle 0=v_{0}\sin(\theta )-gt_{h}}$.

Time to reach the maximum height(h):

${\displaystyle t_{h}={\frac {v_{0}\sin(\theta )}{g}}}$.

For the vertical displacement of the maximum height of projectile:

${\displaystyle h=v_{0}t_{h}\sin(\theta )-{\frac {1}{2}}gt_{h}^{2}}$
${\displaystyle h={\frac {v_{0}^{2}\sin ^{2}(\theta )}{2g}}}$

The maximum reachable height is obtained for θ=90°:

${\displaystyle h_{\mathrm {max} }={\frac {v_{0}^{2}}{2g}}}$

### Relation between horizontal range and maximum height

The relation between the range d on the horizontal plane and the maximum height h reached at ${\displaystyle {\frac {t_{d}}{2}}}$ is:

${\displaystyle h={\frac {d\tan \theta }{4}}}$
Proof

${\displaystyle h={\frac {v_{0}^{2}\sin ^{2}\theta }{2g}}}$

${\displaystyle d={\frac {v_{0}^{2}\sin 2\theta }{g}}}$
${\displaystyle {\frac {h}{d}}={\frac {v_{0}^{2}\sin ^{2}\theta }{2g}}}$ × ${\displaystyle {\frac {g}{v_{0}^{2}\sin 2\theta }}}$
${\displaystyle {\frac {h}{d}}={\frac {\sin ^{2}\theta }{4\sin \theta \cos \theta }}}$

${\displaystyle h={\frac {d\tan \theta }{4}}}$.

### Maximum distance of projectile

The maximum distance of projectile

The range and the maximum height of the projectile does not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction.. The horizontal range d of the projectile is the horizontal distance it has traveled when it returns to its initial height (${\displaystyle y=0}$).

${\displaystyle 0=v_{0}t_{d}\sin(\theta )-{\frac {1}{2}}gt_{d}^{2}}$.

Time to reach ground:

${\displaystyle t_{d}={\frac {2v_{0}\sin(\theta )}{g}}}$.

From the horizontal displacement the maximum distance of projectile:

${\displaystyle d=v_{0}t_{d}\cos(\theta )}$,

so[note 3]

${\displaystyle d={\frac {v_{0}^{2}}{g}}\sin(2\theta )}$.

Note that d has its maximum value when

${\displaystyle \sin 2\theta =1}$,

which necessarily corresponds to

${\displaystyle 2\theta =90^{\circ }}$,

or

${\displaystyle \theta =45^{\circ }}$.
Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).

The total horizontal distance (d) traveled.

${\displaystyle d={\frac {v\cos \theta }{g}}\left(v\sin \theta +{\sqrt {(v\sin \theta )^{2}+2gy_{0}}}\right)}$

When the surface is flat (initial height of the object is zero), the distance traveled:[3]

${\displaystyle d={\frac {v^{2}\sin(2\theta )}{g}}}$

Thus the maximum distance is obtained if θ is 45 degrees. This distance is:

${\displaystyle d_{\mathrm {max} }={\frac {v^{2}}{g}}}$

### Application of the work energy theorem

According to the work-energy theorem the vertical component of velocity is:

${\displaystyle v_{y}^{2}=(v_{0}\sin \theta )^{2}-2gy}$.

These formulae ignore aerodynamic drag and also assume that the landing area is at uniform height 0.

### Angle of reach

The "angle of reach" is the angle (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v.

${\displaystyle \sin(2\theta )={\frac {gd}{v^{2}}}}$

There are two solutions:

${\displaystyle \theta ={\frac {1}{2}}\arcsin \left({\frac {gd}{v^{2}}}\right)}$ (shallow trajectory)

and

${\displaystyle \theta ={\frac {1}{2}}\arccos \left({\frac {gd}{v^{2}}}\right)}$ (steep trajectory)

### Angle θ required to hit coordinate (x, y)

Vacuum trajectory of a projectile for different launch angles. Launch speed is the same for all angles, 50 m/s if "g" is 10 m/s2.

To hit a target at range x and altitude y when fired from (0,0) and with initial speed v the required angle(s) of launch θ are:

${\displaystyle \tan \theta ={\left({\frac {v^{2}\pm {\sqrt {v^{4}-g(gx^{2}+2yv^{2})}}}{gx}}\right)}}$

The two roots of the equation correspond to the two possible launch angles, so long as they aren't imaginary, in which case the initial speed is not great enough to reach the point (x,y) selected. This formula allows one to find the angle of launch needed without the restriction of ${\displaystyle y=0}$.

One can also ask what launch angle allows the lowest possible launch velocity. This occurs when the two solutions above are equal, implying that the quantity under the square root sign is zero. This requires solving a quadratic equation for ${\displaystyle v^{2}}$, and we find

${\displaystyle v^{2}/g=y+{\sqrt {y^{2}+x^{2}}}.}$

This gives

${\displaystyle \theta =\arctan \left(y/x+{\sqrt {y^{2}/x^{2}+1}}\right).}$

If we denote the angle whose tangent is y/x by α, then

${\displaystyle \tan \theta ={\frac {\sin \alpha +1}{\cos \alpha }}}$
${\displaystyle \tan(\pi /2-\theta )={\frac {\cos \alpha }{\sin \alpha +1}}}$
${\displaystyle \cos ^{2}(\pi /2-\theta )={\frac {1}{2}}(\sin \alpha +1)}$
${\displaystyle 2\cos ^{2}(\pi /2-\theta )-1=\cos(\pi /2-\alpha )}$

This implies

${\displaystyle \theta =\pi /2-{\frac {1}{2}}(\pi /2-\alpha ).}$

In other words, the launch should be at the angle halfway between the target and Zenith (vector opposite to Gravity)

### Total Path Length of the Trajectory

The length of the parabolic arc traced by a projectile L, given that the height of launch and landing is the same and that there is no air resistance, is given by the formula:

${\displaystyle L={\frac {v_{0}^{2}}{2g}}\left(2\sin \theta +\cos ^{2}\theta \cdot \ln {\frac {1+\sin \theta }{1-\sin \theta }}\right)={\frac {v_{0}^{2}}{g}}\left(\sin \theta +\cos ^{2}\theta \cdot \tanh ^{-1}(\sin \theta )\right)}$

where ${\displaystyle v_{0}}$ is the initial velocity, ${\displaystyle \theta }$ is the launch angle and ${\displaystyle g}$ is the acceleration due to gravity as a positive value. The expression can be obtained by evaluating the arc length integral for the height-distance parabola between the bounds initial and final displacements (i.e. between 0 and the horizontal range of the projectile) such that:

${\displaystyle L=\int _{0}^{\mathrm {range} }{\sqrt {1+\left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}}}\,\mathrm {d} x=\int _{0}^{v_{0}^{2}\sin(2\theta )/g}{\sqrt {1+\left(-{\frac {g}{v_{0}^{2}\cos ^{2}\theta }}x+\tan \theta \right)^{2}}}\,\mathrm {d} x}$.

## Trajectory of a projectile with air resistance

Trajectories of a mass thrown at an angle of 70°:
without drag
with Stokes' drag
with Newtonian drag

Air resistance creates a force that (for symmetric projectiles) is always directed against the direction of motion in the surrounding medium and has a magnitude that depends on the absolute speed: ${\displaystyle \mathbf {F_{air}} =-f(v)\cdot \mathbf {\hat {v}} }$. The speed-dependence of the friction force is linear (${\displaystyle f(v)\propto v}$) at very low speeds (Stokes drag) and quadratic (${\displaystyle f(v)\propto v^{2}}$) at larger speeds (Newton drag).[4] The transition between these behaviours is determined by the Reynolds number, which depends on speed, object size and kinematic viscosity of the medium. For Reynolds numbers below about 1000, the dependence is linear, above it becomes quadratic. In air, which has a kinematic viscosity around ${\displaystyle 0.15\,\mathrm {cm^{2}/s} }$, this means that the drag force becomes quadratic in v when the product of speed and diameter is more than about ${\displaystyle 0.015\,\mathrm {m^{2}/s} }$, which is typically the case for projectiles.

• Stokes drag: ${\displaystyle \mathbf {F_{air}} =-k_{\mathrm {Stokes} }\cdot \mathbf {v} \qquad }$ (for ${\displaystyle Re\lesssim 1000}$)
• Newton drag: ${\displaystyle \mathbf {F_{air}} =-k\,|\mathbf {v} |\cdot \mathbf {v} \qquad }$ (for ${\displaystyle Re\gtrsim 1000}$)
Free body diagram of a body on which only gravity and air resistance acts

The free body diagram on the right is for a projectile that experiences air resistance and the effects of gravity. Here, air resistance is assumed to be in the direction opposite of the projectile's velocity: ${\displaystyle \mathbf {F_{\mathrm {air} }} =-f(v)\cdot \mathbf {\hat {v}} }$

### Trajectory of a projectile with Stokes drag

Stokes drag, where ${\displaystyle \mathbf {F_{air}} \propto \mathbf {v} }$, only applies at very low speed in air, and is thus not the typical case for projectiles. However, the linear dependence of ${\displaystyle F_{\mathrm {air} }}$ on ${\displaystyle v}$ causes a very simple differential equation of motion

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\begin{pmatrix}v_{x}\\v_{y}\end{pmatrix}}={\begin{pmatrix}-\mu \,v_{x}\\-g-\mu \,v_{y}\end{pmatrix}}}$

in which the two cartesian components become completely independent, and thus easier to solve.[5] Here, ${\displaystyle v_{0}}$,${\displaystyle v_{x}}$ and ${\displaystyle v_{y}}$ will be used to denote the initial velocity, the velocity along the direction of x and the velocity along the direction of y, respectively. The mass of the projectile will be denoted by m, and ${\displaystyle \mu :=k/m}$. For the derivation only the case where ${\displaystyle 0^{o}\leq \theta \leq 180^{o}}$ is considered. Again, the projectile is fired from the origin (0,0).

Derivation of horizontal position

The relationships that represent the motion of the particle are derived by Newton's Second Law, both in the x and y directions. In the x direction ${\displaystyle \Sigma F=-kv_{x}=ma_{x}}$ and in the y direction ${\displaystyle \Sigma F=-kv_{y}-mg=ma_{y}}$.

This implies that:

${\displaystyle a_{x}=-\mu v_{x}={\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}}$ (1),

and

${\displaystyle a_{y}=-\mu v_{y}-g={\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}}$ (2)

Solving (1) is an elementary differential equation, thus the steps leading to a unique solution for vx and, subsequently, x will not be enumerated. Given the initial conditions ${\displaystyle v_{x}=v_{x0}}$ (where vx0 is understood to be the x component of the initial velocity) and ${\displaystyle x=0}$ for ${\displaystyle t=0}$:

${\displaystyle v_{x}=v_{x0}e^{-\mu t}}$ (1a)

${\displaystyle x(t)={\frac {v_{x0}}{\mu }}\left(1-e^{-\mu t}\right)}$ (1b)
Derivation of vertical position

While (1) is solved much in the same way, (2) is of distinct interest because of its non-homogeneous nature. Hence, we will be extensively solving (2). Note that in this case the initial conditions are used ${\displaystyle v_{y}=v_{y0}}$ and ${\displaystyle y=0}$ when ${\displaystyle t=0}$.

${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}=-\mu v_{y}-g}$ (2)

${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}+\mu v_{y}=-g}$ (2a)

This first order, linear, non-homogeneous differential equation may be solved a number of ways; however, in this instance, it will be quicker to approach the solution via an integrating factor ${\displaystyle e^{\int \mu \,\mathrm {d} t}}$.

${\displaystyle e^{\mu t}({\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}+\mu v_{y})=e^{\mu t}(-g)}$ (2c)

${\displaystyle (e^{\mu t}v_{y})^{\prime }=e^{\mu t}(-g)}$ (2d)

${\displaystyle \int {(e^{\mu t}v_{y})^{\prime }\,\mathrm {d} t}=e^{\mu t}v_{y}=\int {e^{\mu t}(-g)\,\mathrm {d} t}}$ (2e)

${\displaystyle e^{\mu t}v_{y}={\frac {1}{\mu }}e^{\mu t}(-g)+C}$(2f)

${\displaystyle v_{y}={\frac {-g}{\mu }}+Ce^{-\mu t}}$ (2g)

And by integration we find:

${\displaystyle y=-{\frac {g}{\mu }}t-{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}+C}$ (3)

Solving for our initial conditions:

${\displaystyle v_{y}(t)=-{\frac {g}{\mu }}+(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}}$ (2h)

${\displaystyle y(t)=-{\frac {g}{\mu }}t-{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}+{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})}$ (3a)

With a bit of algebra to simplify (3a):

${\displaystyle y(t)=-{\frac {g}{\mu }}t+{\frac {1}{\mu }}\left(v_{y0}+{\frac {g}{\mu }}\right)\left(1-e^{-\mu t}\right)}$ (3b)
Derivation of the time of flight

The total time of the journey in the presence of air resistance (more specifically, when ${\displaystyle F_{air}=-kv}$) can be calculated by the same strategy as above, namely, we solve the equation ${\displaystyle y(t)=0}$. While in the case of zero air resistance this equation can be solved elementarily, here we shall need the Lambert W function. The equation ${\displaystyle y(t)=-{\frac {g}{\mu }}t+{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})(1-e^{-\mu t})=0}$ is of the form ${\displaystyle c_{1}t+c_{2}+c_{3}e^{c_{4}t}=0}$, and such an equation can be transformed into an equation solvable by the ${\displaystyle W}$ function (see an example of such a transformation here). Some algebra shows that the total time of flight, in closed form, is given as[6]

${\displaystyle t={\frac {1}{\mu }}\left(1+{\frac {\mu }{g}}v_{y0}+W\left(-\left(1+{\frac {\mu }{g}}v_{y0}\right)e^{-\left(1+{\frac {\mu }{g}}v_{y0}\right)}\right)\right)}$.

### Trajectory of a projectile with Newton drag

Trajectories of a skydiver in air with Newton drag

The most typical case of air resistance, for the case of Reynolds numbers above about 1000 is Newton drag with a drag force proportional to the speed squared, ${\displaystyle F_{\mathrm {air} }=-kv^{2}}$. In air, which has a kinematic viscosity around ${\displaystyle 0.15\,\mathrm {cm^{2}/s} }$, this means that the product of speed and diameter must be more than about ${\displaystyle 0.015\,\mathrm {m^{2}/s} }$.

Unfortunately, the equations of motion can not be easily solved analytically for this case. Therefore, a numerical solution will be examined.

${\displaystyle \mathbf {F_{D}} =-{\tfrac {1}{2}}c\rho A\,v\,\mathbf {v} }$
Where:

#### Special cases

Even though the general case of a projectile with Newton drag cannot be solved analytically, some special cases can. Here we denote the terminal velocity in free-fall as ${\displaystyle v_{\infty }={\sqrt {g/\mu }}}$ and the characteristic settling time constant ${\displaystyle t_{f}=1/{\sqrt {g\mu }}}$.

• Near-horizontal motion: In case the motion is almost horizontal, ${\displaystyle |v_{x}|\gg |v_{y}|}$, such as a flying bullet, the vertical velocity component has very little influence on the horizontal motion. In this case:[7]
${\displaystyle {\dot {v}}_{x}(t)=-\mu \,v_{x}^{2}(t)}$
${\displaystyle v_{x}(t)={\frac {1}{1/v_{x,0}+\mu \,t}}}$
${\displaystyle x(t)={\frac {1}{\mu }}\ln(1+\mu \,v_{x,0}\cdot t)}$
The same pattern applies for motion with friction along a line in any direction, when gravity is negligible. It also applies when vertical motion is prevented, such as for a moving car with its engine off.
• Vertical motion upward:[7]
${\displaystyle {\dot {v}}_{y}(t)=-g-\mu \,v_{y}^{2}(t)}$
${\displaystyle v_{y}(t)=v_{\infty }\tan {\frac {t_{\mathrm {peak} }-t}{t_{f}}}}$
${\displaystyle y(t)=y_{\mathrm {peak} }+{\frac {1}{\mu }}\ln \left(\cos {\frac {t_{\mathrm {peak} }-t}{t_{f}}}\right)}$
A projectile can not rise longer than ${\displaystyle t_{\mathrm {rise} }={\frac {\pi }{2}}t_{f}}$ vertically before it reaches the peak.
• Vertical motion downward:[7]
${\displaystyle {\dot {v}}_{y}(t)=-g+\mu \,v_{y}^{2}(t)}$
${\displaystyle v_{y}(t)=-v_{\infty }\tanh {\frac {t-t_{\mathrm {peak} }}{t_{f}}}}$
${\displaystyle y(t)=y_{\mathrm {peak} }-{\frac {1}{\mu }}\ln \left(\cosh {\frac {t-t_{\mathrm {peak} }}{t_{f}}}\right)}$
After a time ${\displaystyle t_{f}}$, the projectile reaches almost terminal velocity ${\displaystyle -v_{\infty }}$.

#### Integral expressions

The approach will be to formulate integral expressions that can be evaluated numerically. All variables will then be expressed in terms of a parameter ${\displaystyle Q}$.

Derivation of integral expressions

A projectile of mass m is launched from a point ${\displaystyle (x_{0},y_{0})}$, with an initial velocity ${\displaystyle \mathbf {v_{0}} }$ in an initial direction that makes an angle ${\displaystyle \Psi _{0}}$ with the horizontal. It experiences air resistance that is given by ${\displaystyle F_{air}=-kv^{2}}$ that acts tangentially to the path of travel at any point.

Newton's second law of motion is ${\displaystyle \mathbf {F} =m\mathbf {a} }$. Applying this in the x-direction yields;

${\displaystyle F_{x}=-kv^{2}\cos \Psi =m{\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}}$

(1)

Where, ${\displaystyle v={\sqrt {v_{x}^{2}+v_{y}^{2}}}}$, ${\displaystyle v_{x}}$ and ${\displaystyle v_{y}}$ are the horizontal and vertical components of the velocity ${\displaystyle v}$ respectively.

Let ${\displaystyle \mu ={\frac {k}{m}}}$, ${\displaystyle {\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}={\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}}$, and ${\displaystyle \cos \Psi ={\frac {v_{x}}{v}}}$. Equation (1) now becomes;

${\displaystyle {\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}=-\mu v_{x}{\sqrt {v_{x}^{2}+v_{y}^{2}}}}$

(A)

In the y-direction;

${\displaystyle F_{y}=-kv^{2}\sin \Psi -mg=m{\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}}}$

(2)

Again let, ${\displaystyle \mu ={\frac {k}{m}}}$, ${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}={\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}}}$, and ${\displaystyle \sin \Psi ={\frac {v_{y}}{v}}}$. Equation (2) is now;

${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}=-\mu v_{y}{\sqrt {v_{x}^{2}+v_{y}^{2}}}-g}$

(B)

Knowing that ${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} v_{x}}}={\frac {{\Bigl (}{\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}{\Bigr )}}{{\Bigl (}{\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}{\Bigr )}}}}$ we may divide equation (B) by equation (A) to get;

${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} v_{x}}}={\frac {v_{y}}{v_{x}}}+{\frac {g}{\mu v_{x}{\sqrt {v_{x}^{2}+v_{y}^{2}}}}}}$

(C)

Introduce a quantity ${\displaystyle q}$ such that ${\displaystyle v_{y}=qv_{x}}$, then;

${\displaystyle {\frac {\mathrm {d} v_{y}}{\mathrm {d} v_{x}}}={\frac {\mathrm {d} }{\mathrm {d} v_{x}}}(qv_{x})=q+v_{x}{\frac {\mathrm {d} q}{\mathrm {d} u}}}$

(D)

From equations (C) and (D), observe that; ${\displaystyle {\frac {v_{y}}{v_{x}}}+{\frac {g}{\mu v_{x}{\sqrt {v_{x}^{2}+v_{y}^{2}}}}}=q+v_{x}{\frac {\mathrm {d} q}{\mathrm {d} v_{x}}}}$

Hence, ${\displaystyle v_{x}{\frac {\mathrm {d} q}{\mathrm {d} v_{x}}}={\frac {g}{\mu v_{x}{\sqrt {v_{x}^{2}+v_{y}^{2}}}}}}$ which may be re-written as; ${\displaystyle v_{x}{\frac {\mathrm {d} q}{\mathrm {d} v_{x}}}={\frac {g}{\mu v_{x}^{2}{\sqrt {1+q^{2}}}}}}$

Separate variables and integrate as;

${\displaystyle {\frac {g}{\mu }}\int v_{x}^{-3}\,\mathrm {d} v_{x}=\int {\sqrt {1+q^{2}}}\,\mathrm {d} q}$

(E)

The left-hand side of equation (E) is ${\displaystyle \int v_{x}^{-3}\,\mathrm {d} v_{x}=-{\frac {1}{2v_{x}^{2}}}}$

For the right-hand side, let ${\displaystyle q=\sinh Q}$, such that ${\displaystyle 1+q^{2}=1+\sinh ^{2}Q=\cosh ^{2}Q}$ and, ${\displaystyle \mathrm {d} q=\cosh Q\,\mathrm {d} Q}$

Thus ${\displaystyle \int {\sqrt {1+q^{2}}}\,\mathrm {d} q=\int \cosh ^{2}Q\,\mathrm {d} Q}$. Also ${\displaystyle \cosh ^{2}Q={\frac {1}{2}}(1+\cosh {2Q})}$

Hence; ${\displaystyle \int \cosh ^{2}Q\,\mathrm {d} Q={\frac {1}{2}}{\biggl \{}Q+{\frac {1}{2}}\sinh {2Q}{\biggr \}}}$

Equation (E) is now; ${\displaystyle -{\frac {g}{2\mu v_{x}^{2}}}={\frac {1}{2}}{\biggl \{}Q+{\frac {1}{2}}\sinh {2Q}{\biggr \}}-{\tilde {B}}}$

From which;

${\displaystyle v_{x}={\sqrt {\frac {g}{\mu }}}{\frac {1}{\sqrt {{\biggl \{}B-{\Bigl (}Q+{\frac {1}{2}}\sinh {2Q}{\Bigr )}{\biggr \}}}}}}$

Since ${\displaystyle v_{y}=v_{x}q=v_{x}\sinh {Q}}$

${\displaystyle v_{y}={\sqrt {\frac {g}{\mu }}}{\frac {\sinh {Q}}{\sqrt {{\biggl \{}B-{\Bigl (}Q+{\frac {1}{2}}\sinh {2Q}{\Bigr )}{\biggr \}}}}}}$

Denote ${\displaystyle \lambda =B-{\Bigl (}Q+{\frac {1}{2}}\sinh {2Q}{\Bigr )}}$, such that;

${\displaystyle v_{x}={\sqrt {\frac {g}{\mu }}}{\frac {1}{\sqrt {\lambda }}}}$

(F)

${\displaystyle v_{y}={\sqrt {\frac {g}{\mu }}}{\frac {\sinh {Q}}{\sqrt {\lambda }}}}$

(G)

At the beginning of the motion, ${\displaystyle t=0}$ and ${\displaystyle v_{0}^{2}=v_{x,0}^{2}+v_{y,0}^{2}}$

Hence; ${\displaystyle \tan {\Psi _{0}}={\frac {v_{y,0}}{v_{x,0}}}=q_{0}=\sinh {Q_{0}}}$, such that; ${\displaystyle B={\frac {g}{\mu v_{x,0}^{2}}}+{\biggl \{}Q_{0}+{\frac {1}{2}}\sinh {2Q_{0}}{\biggr \}}}$

As the motion proceeds, ${\displaystyle v_{x}\rightarrow 0}$ and ${\displaystyle v_{y}\rightarrow (a\,negative\,constant)}$, i.e., ${\displaystyle q\rightarrow -\infty }$, and ${\displaystyle Q\rightarrow -\infty }$

This means that, ${\displaystyle \lambda \to +\infty }$ and ${\displaystyle {\frac {1}{\sqrt {\lambda }}}\to 0}$

Hence; ${\displaystyle \lim _{Q\to -\infty }{\frac {\sinh {Q}}{\sqrt {\lambda }}}=-1}$

In equations (F) and (G), observe that;

As ${\displaystyle v_{x}\to 0}$, ${\displaystyle v_{y}\to {\sqrt {\frac {g}{\mu }}}}$

When a state of dynamic equilibrium is attained under vertical free fall, the opposing forces of gravity and drag are equalized, i.e., ${\displaystyle kv_{y,t}^{2}=mg}$

${\displaystyle v_{y,t}={\sqrt {\frac {g}{\mu }}}}$

(H)

In equation (A), substitutions for ${\displaystyle v_{x}}$ and ${\displaystyle v_{y}}$ from equations (F) and (G) yields;

${\displaystyle {\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}=-{\frac {g}{\lambda }}\cosh {Q}}$

Also; ${\displaystyle {\frac {\mathrm {d} v_{x}}{\mathrm {d} Q}}={\frac {1}{2\lambda ^{3/2}}}{\sqrt {\frac {g}{\mu }}}(1+\cosh {2Q})}$

Knowing that; ${\displaystyle {\frac {\mathrm {d} Q}{\mathrm {d} t}}={\frac {{\Bigl (}{\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}{\Bigr )}}{{\Bigl (}{\frac {\mathrm {d} v_{x}}{\mathrm {d} Q}}{\Bigr )}}}}$, we may write

${\displaystyle {\frac {\mathrm {d} t}{\mathrm {d} Q}}=-{\frac {\cosh {Q}}{{\sqrt {g\mu }}{\sqrt {\lambda }}}}}$
${\displaystyle t(Q)=-{\frac {1}{\sqrt {g\mu }}}\int _{Q_{0}}^{Q}{\frac {\cosh {\tilde {Q}}}{\sqrt {\lambda }}}\,\mathrm {d} {\tilde {Q}}}$

(I)

Also; ${\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} Q}}={\frac {{\Bigl (}{\frac {\mathrm {d} x}{\mathrm {d} t}}{\Bigr )}}{{\Bigl (}{\frac {\mathrm {d} Q}{\mathrm {d} t}}{\Bigr )}}}=-{\frac {1}{\mu }}{\frac {\cosh {Q}}{\lambda }}}$

${\displaystyle x(Q)=x_{0}-{\frac {1}{\mu }}\int _{Q_{0}}^{Q}{\frac {\cosh {\tilde {Q}}}{\lambda }}\,\mathrm {d} {\tilde {Q}}}$

(J)

And; ${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} Q}}={\frac {{\Bigl (}{\frac {\mathrm {d} y}{\mathrm {d} t}}{\Bigr )}}{{\Bigl (}{\frac {\mathrm {d} Q}{\mathrm {d} t}}{\Bigr )}}}=-{\frac {1}{2\mu }}{\frac {\sinh {2Q}}{\lambda }}}$

${\displaystyle y(Q)=y_{0}-{\frac {1}{2\mu }}\int _{Q_{0}}^{Q}{\frac {\sinh {2{\tilde {Q}}}}{\lambda }}\,\mathrm {d} {\tilde {Q}}}$

(K)

Determine the time of flight ${\displaystyle T}$ by setting ${\displaystyle y}$ to ${\displaystyle 0}$ in equation (K) above. Solve for the value of the variable ${\displaystyle Q}$.

${\displaystyle y_{0}-{\frac {1}{2\mu }}\int _{Q_{0}}^{Q_{T}}{\frac {\sinh {2{\tilde {Q}}}}{\lambda }}\,\mathrm {d} {\tilde {Q}}=0}$

(L)

Equation (I) with ${\displaystyle Q_{T}}$ substituted for ${\displaystyle Q}$ gives;

${\displaystyle T=-{\frac {1}{\sqrt {g\mu }}}\int _{Q_{0}}^{Q_{T}}{\frac {\cosh {\tilde {Q}}}{\sqrt {\lambda }}}\,\mathrm {d} {\tilde {Q}}}$

(M)

Equation (J) gives the horizontal range ${\displaystyle R}$ as;

${\displaystyle R=x_{0}-{\frac {1}{\mu }}\int _{Q_{0}}^{Q_{T}}{\frac {\cosh {\tilde {Q}}}{\lambda }}\,\mathrm {d} {\tilde {Q}}}$

(N)

At the highest point of the projectile path ${\displaystyle \Psi =0}$, and ${\displaystyle Q=0}$, giving the maximum height ${\displaystyle H}$ from equation (K) as;

${\displaystyle H=y_{0}-{\frac {1}{2\mu }}\int _{Q_{0}}^{0}{\frac {\sinh {2{\tilde {Q}}}}{\lambda }}\,\mathrm {d} {\tilde {Q}}}$

(O)

#### Numerical solution

A projectile motion with drag can be computed generically by numerical integration of the ordinary differential equation, for instance by applying a reduction to a first-order system. The equation to be solved is

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\begin{pmatrix}x\\y\\v_{x}\\v_{y}\end{pmatrix}}={\begin{pmatrix}v_{x}\\v_{y}\\-\mu \,v_{x}{\sqrt {v_{x}^{2}+v_{y}^{2}}}\\-g-\mu \,v_{y}{\sqrt {v_{x}^{2}+v_{y}^{2}}}\end{pmatrix}}}$.

This approach also allows to add the effects of speed-dependent drag coefficient, altitude-dependent air density and position-dependent gravity field.

## Lofted trajectory

Lofted trajectories of North Korean missiles Hwasong-14 and Hwasong-15

A special case of a ballistic trajectory for a rocket is a lofted trajectory, a trajectory with an apogee greater than the minimum-energy trajectory to the same range. In other words, the rocket travels higher and by doing so it uses more energy to get to the same landing point. This may be done for various reasons such as increasing distance to the horizon to give greater viewing/communication range or for changing the angle with which a missile will impact on landing. Lofted trajectories are sometimes used in both missile rocketry and in spaceflight.[8]

## Projectile motion on a planetary scale

Projectile trajectory around a planet, compared to the motion in a uniform field

When a projectile without air resistance travels a range that is significant compared to the earth's radius (above ≈100 km), the curvature of the earth and the non-uniform gravitational field have to be considered. This is for example the case with spacecraft or intercontinental projectiles. The trajectory then generalizes from a parabola to a Kepler-ellipse with one focus at the center of the earth. The projectile motion then follows Kepler's laws of planetary motion.

The trajectories' parameters have to be adapted from the values of a uniform gravity field stated above. The earth radius is taken as R, and g as the standard surface gravity. Let ${\displaystyle {\tilde {v}}:=v/{\sqrt {Rg}}}$ the launch velocity relative to the first cosmic velocity.

Total range d between launch and impact:

${\displaystyle d={\frac {v^{2}\sin(2\theta )}{g}}{\Big /}{\sqrt {1-\left(2-{\tilde {v}}^{2}\right){\tilde {v}}^{2}\cos ^{2}\theta }}}$

Maximum range of a projectile for optimum launch angle (${\displaystyle \theta ={\tfrac {1}{2}}\arccos \left({\tilde {v}}^{2}/(2-{\tilde {v}}^{2})\right)}$):

${\displaystyle d_{\mathrm {max} }={\frac {v^{2}}{g}}{\big /}\left(1-{\tfrac {1}{2}}{\tilde {v}}^{2}\right)}$       with ${\displaystyle v<{\sqrt {Rg}}}$, the first cosmic velocity

Maximum height of a projectile above the planetary surface:

${\displaystyle h={\frac {v^{2}\sin ^{2}\theta }{g}}{\Big /}\left(1-{\tilde {v}}^{2}+{\sqrt {1-\left(2-{\tilde {v}}^{2}\right){\tilde {v}}^{2}\cos ^{2}\theta }}\right)}$

Maximum height of a projectile for vertical launch (${\displaystyle \theta =90^{\circ }}$):

${\displaystyle h_{\mathrm {max} }={\frac {v^{2}}{2g}}{\big /}\left(1-{\tfrac {1}{2}}{\tilde {v}}^{2}\right)}$       with ${\displaystyle v<{\sqrt {2Rg}}}$, the second cosmic velocity

Time of flight:

${\displaystyle t={\frac {2v\sin \theta }{g}}\cdot {\frac {1}{2-{\tilde {v}}^{2}}}\left(1+{\frac {1}{{\sqrt {2-{\tilde {v}}^{2}}}\,{\tilde {v}}\sin \theta }}\arcsin {\frac {{\sqrt {2-{\tilde {v}}^{2}}}\,{\tilde {v}}\sin \theta }{\sqrt {1-\left(2-{\tilde {v}}^{2}\right){\tilde {v}}^{2}\cos ^{2}\theta }}}\right)}$

## Notes

1. ^ The g is the acceleration due to gravity. (${\displaystyle 9.81\,\mathrm {m/s^{2}} }$ near the surface of the Earth).
2. ^ decreasing when the object goes upward, and increasing when it goes downward
3. ^ ${\displaystyle 2\cdot \sin(\alpha )\cdot \cos(\alpha )=\sin(2\alpha )}$

## References

1. ^ Galileo Galilei, Two New Sciences, Leiden, 1638, p.249
2. ^ Nolte, David D., Galileo Unbound (Oxford University Press, 2018) pp. 39-63.
3. ^ Tatum (2019). Classical Mechanics (PDF). pp. ch. 7.
4. ^ Stephen T. Thornton; Jerry B. Marion (2007). Classical Dynamics of Particles and Systems. Brooks/Cole. p. 59. ISBN 978-0-495-55610-7.
5. ^ Atam P. Arya; Atam Parkash Arya (September 1997). Introduction to Classical Mechanics. Prentice Hall Internat. p. 227. ISBN 978-0-13-906686-3.
6. ^ Rginald Cristian, Bernardo; Jose Perico, Esguerra; Jazmine Day, Vallejos; Jeff Jerard, Canda (2015). "Wind-influenced projectile motion". European Journal of Physics. 36 (2). doi:10.1088/0143-0807/36/2/025016.
7. ^ a b c Walter Greiner (2004). Classical Mechanics: Point Particles and Relativity. Springer Science & Business Media. p. 181. ISBN 0-387-95586-0.
8. ^