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Projectile motion

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Motion of launched objects due to gravity

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Components of initial velocity of parabolic throwing

Ballistic trajectories are parabolic if gravity is homogeneous and elliptic if it is radial.

In physics, projectile motion describes the motion of an object that is launched into the air and moves under the influence of gravity alone, with air resistance neglected. In this idealized model, the object follows a parabolic path determined by its initial velocity and the constant acceleration due to gravity. The motion can be decomposed into horizontal and vertical components: the horizontal motion occurs at a constant velocity, while the vertical motion experiences uniform acceleration.

This framework, which lies at the heart of classical mechanics, is fundamental to a wide range of applications—from engineering and ballistics to sports science and natural phenomena.

Galileo Galilei showed that the trajectory of a given projectile isparabolic, but the path may also bestraight in the special case when the object is thrown directly upward or downward. The study of such motions is calledballistics, and such a trajectory is described asballistic. The only force of mathematical significance that is actively exerted on the object is gravity, which acts downward, thus imparting to the object a downwardacceleration towards Earth'scenter of mass. Due to the object'sinertia, no external force is needed to maintain the horizontal velocitycomponent of the object's motion.

Taking other forces into account, such asaerodynamic drag or internal propulsion (such as in arocket), requires additional analysis. Aballistic missile is amissile onlyguided during the relatively brief initialpowered phase of flight, and whose remaining course is governed by the laws ofclassical mechanics.

Ballistics (from Ancient Greek βάλλεινbállein 'to throw') is the science ofdynamics that deals with the flight, behavior and effects of projectiles, especiallybullets,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, the launch angle and a gravitational acceleration assumed constant. Practical solutions of a ballistics problem often require considerations of air resistance, cross winds, target motion, acceleration due to gravity varying with height, and in such problems aslaunching a rocket from one point on the Earth to another, the horizon's distancevs curvature R of the Earth (its local speed of rotation ${\textstyle v(lat)=\omega R(lat)}$ ). Detailed mathematical solutions of practical problems typically do not haveclosed-form solutions, and therefore requirenumerical methods to address.

Kinematic quantities

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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 ofcompound motion established byGalileo 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 as ${\textstyle g(y)=g_{0}/(1+y/R)^{2}}$ and direction (faraway targets) with latitude/longitude along the trajectory. This causes anelliptic 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 ablack hole of equal mass, it would become obvious that the ballistic trajectory is part of an ellipticorbit around that "black hole", and not a parabola that extends to infinity. At higher speeds the trajectory can also be circular (cosmonautics atLEO?,geostationary satellites at 5 ${\textstyle {\frac {5}{6}}}$ R), parabolic orhyperbolic (unless distorted by other objects like the Moon or the Sun).

In this article ahomogeneous gravitational acceleration ${\textstyle (g=g_{0})}$ is assumed.

Acceleration

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

a_{x}=0

a_{y}=-g

*The y acceleration can also be referred to as the force of the earth ${\textstyle (-F_{g}/m)}$ on the object(s) of interest.

Velocity

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

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

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

v_{0x}=v_{0}\cos(\theta )

v_{0y}=v_{0}\sin(\theta )

The horizontal component of thevelocity 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 thex andy directions can be integrated to solve for the components of velocity at any timet, as follows:

v_{x}=v_{0}\cos(\theta )

v_{y}=v_{0}\sin(\theta )-gt

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

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

Displacement

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

x=v_{0}t\cos(\theta )

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

The magnitude of the displacement is:

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

Consider the equations,

x=v_{0}t\cos(\theta )

and

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

.^[3]

Ift is eliminated between these two equations the following equation is obtained:

y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}=\tan \theta \cdot x\left(1-{\frac {x}{R}}\right).

Here R is therange of a projectile.

Sinceg,θ, andv₀ are constants, the above equation is of the form

y=ax+bx^{2}

in whicha andb 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 afore-mentioned parabolic equation:

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

Displacement in polar coordinates

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

r(\phi )={\frac {2v_{0}^{2}\cos ^{2}\theta }{|g|}}\left(\tan \theta \sec \phi -\tan \phi \sec \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 $0\leq \phi \leq \pi$ . This expression can be obtained by transforming the Cartesian equation as stated above by $y=r\sin \phi$ and $x=r\cos \phi$ .

Properties of the trajectory

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Time of flight or total time of the whole journey

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

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

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

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

v_{0}\sin(\theta )={\frac {1}{2}}gt

t={\frac {2v_{0}\sin(\theta )}{g}}

Note that we have neglected air resistance on the projectile.

If the starting point is at heighty₀ with respect to the point of impact, the time of flight is:

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 (y₀ is 0) to

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

{\frac {2v\sin {(45^{\circ })}}{|g|}}={\frac {{\sqrt {2}}v}{|g|}}

ifθ equals 45°.

Time of flight to the target's position

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As shown above in theDisplacement section, the horizontal and vertical velocity of a projectile are independent of each other.

Because of this, we can find the time to reach a target using the displacement formula for the horizontal velocity:

$x=v_{0}t\cos(\theta )$

${\frac {x}{t}}=v_{0}\cos(\theta )$

$t={\frac {x}{v_{0}\cos(\theta )}}$

This equation will give the total timet the projectile must travel for to reach the target's horizontal displacement, neglecting air resistance.

Maximum height of projectile

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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 $v_{y}=0$ , that is,

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

Time to reach the maximum height(h):

{\textstyle t_{h}={\frac {v_{0}\sin(\theta )}{|g|}}}

For the vertical displacement of the maximum height of the projectile:

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

h={\frac {v_{0}^{2}\sin ^{2}(\theta )}{2|g|}}

The maximum reachable height is obtained forθ=90°:

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

If the projectile's position (x,y) and launch angle (θ) are known, the maximum height can be found by solving for h in the following equation:

h={\frac {(x\tan \theta )^{2}}{4(x\tan \theta -y)}}.

Angle ofelevation (φ) at the maximum height is given by:

\phi =\arctan {\tan \theta  \over 2}

Relation between horizontal range and maximum height

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The relation between the ranged on the horizontal plane and the maximum heighth reached at ${\frac {t_{d}}{2}}$ is:

h={\frac {d\tan \theta }{4}}

Proof
$h={\frac {v_{0}^{2}\sin ^{2}\theta }{2\|g\|}}$ $d={\frac {v_{0}^{2}\sin 2\theta }{\|g\|}}$ ${\frac {h}{d}}={\frac {v_{0}^{2}\sin ^{2}\theta }{2\|g\|}}$ × ${\frac {g}{v_{0}^{2}\sin 2\theta }}$ ${\frac {h}{d}}={\frac {\sin ^{2}\theta }{4\sin \theta \cos \theta }}$ $h={\frac {d\tan \theta }{4}}$ . If $h=R$ $\theta =\arctan(4)\approx 76.0^{\circ }$

Proof

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

d={\frac {v_{0}^{2}\sin 2\theta }{|g|}}

{\frac {h}{d}}={\frac {v_{0}^{2}\sin ^{2}\theta }{2|g|}}

{\frac {g}{v_{0}^{2}\sin 2\theta }}

{\frac {h}{d}}={\frac {\sin ^{2}\theta }{4\sin \theta \cos \theta }}

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

If $h=R$

$\theta =\arctan(4)\approx 76.0^{\circ }$

Maximum distance of projectile

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Main article:Range of a projectile

The range and the maximum height of the projectile do 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 ranged of the projectile is the horizontal distance it has traveled when it returns to its initial height ( ${\textstyle y=0}$ ).

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

Time to reach ground:

t_{d}={\frac {2v_{0}\sin(\theta )}{|g|}}

From the horizontal displacement the maximum distance of the projectile:

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

so^{[note 3]}

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

Note thatd has its maximum value when

\sin(2\theta )=1,

which necessarily corresponds to ${\textstyle 2\theta =90^{\circ }}$ , or ${\textstyle \theta =45^{\circ }}$ .

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 andH = highest point of trajectory (indicated with arrows).

The total horizontal distance(d) traveled.

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:^[4]

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

Thus the maximum distance is obtained ifθ is 45 degrees. This distance is:

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

Application of the work energy theorem

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

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

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

Angle of reach

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

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

There are two solutions:

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

(shallow trajectory)

and because $\sin(2\theta )=\cos(2\theta -90^{\circ })$ ,

\theta =45^{\circ }+{\frac {1}{2}}\arccos \left({\frac {gd}{v^{2}}}\right)

(steep trajectory)

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

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

To hit a target at rangex and altitudey when fired from (0,0) and with initial speedv, the required angle(s) of launchθ are:

\theta =\arctan {\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 ${\textstyle y=0}$ .

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, tan θ = v²/gx, requires solving a quadratic equation for $v^{2}$ ,^[5] and we find

^{$v^{2}/g=(2gy\pm g{\sqrt {4y^{2}+4(1x)^{2}}})/2g=y\pm {\sqrt {y^{2}+x^{2}}}.$}

This gives

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

If we denote the angle whose tangent isy/x byα,^[6] then

\tan \theta ={\frac {\sin \alpha +1}{\cos \alpha }},

its reciprocal:

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

^[7]

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

2\cos ^{2}(\pi /2-\theta )-1=\cos(\pi /2-\alpha )

This implies

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

In other words, the launch should be at the angle ${\textstyle (\pi /2+\alpha )/2}$ halfway between the target andzenith (vector opposite to gravity).

Total Path Length of the Trajectory

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

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

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(\tan \theta -{g \over {v_{0}^{2}\cos ^{2}\theta }}x\right)^{2}}}\,\mathrm {d} x.

If the time-of-flight ist,

L=\int _{0}^{t}{\sqrt {v_{x}^{2}+v_{y}^{2}}}\,\mathrm {d} t=\int _{0}^{2v_{0}\sin \theta /g}{\sqrt {(gt)^{2}-2gv_{0}\sin \theta t+v_{0}^{2}}}\,\mathrm {d} t.

Trajectory of a projectile with air resistance

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Trajectories of a mass thrown at an angle of 70°:
withoutdrag (a parabole)
withStokes' drag
withNewtonian 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: $\mathbf {F_{air}} =-f(v)\cdot \mathbf {\hat {v}}$ . The speed-dependence of the friction force is linear ( $f(v)\propto v$ ) at very low speeds (Stokes drag) and quadratic ( $f(v)\propto v^{2}$ ) at large speeds (Newton drag).^[8] The transition between these behaviours is determined by theReynolds number, which depends on object speed and size, density ${\textstyle \rho }$ anddynamic viscosity ${\textstyle \eta }$ of the medium. For Reynolds numbers below about 1 the dependence is linear, above 1000 (turbulent flow) it becomes quadratic. In air, which has akinematic viscosity $\eta /\rho$ around 0.15 cm²/s, this means that the drag force becomes quadratic inv when the product of object speed and diameter is more than about 0.015 m²/s, which is typically the case for projectiles.

Stokes drag: $\mathbf {F_{air}} =-k_{\mathrm {Stokes} }\cdot \mathbf {v} \qquad$ (for ${\textstyle Re\lesssim 1}$ )
Newton drag: $\mathbf {F_{air}} =-k\,|\mathbf {v} |\cdot \mathbf {v} \qquad$ (for ${\textstyle Re\gtrsim 1000}$ )

Free body diagram of a body on which only gravity and air resistance act.

Thefree 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: $\mathbf {F_{\mathrm {air} }} =-f(v)\cdot \mathbf {\hat {v}}$

Trajectory of a projectile with Stokes drag

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

{\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 2 cartesian components become completely independent, and it is thus easier to solve.^[9] Here, $v_{0}$ , $v_{x}$ and $v_{y}$ will be used to denote the initial velocity, the velocity along the direction ofx and the velocity along the direction ofy, respectively. The mass of the projectile will be denoted bym, and $\mu :=k/m$ . For the derivation only the case where ${\textstyle 0^{o}\leq \theta \leq 180^{o}}$ 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 byNewton's second law, both in the x and y directions. In the x direction $\Sigma F=-kv_{x}=ma_{x}$ and in the y direction $\Sigma F=-kv_{y}-mg=ma_{y}$ . This implies that: $a_{x}=-\mu v_{x}={\frac {\mathrm {d} v_{x}}{\mathrm {d} t}}$ (1), and $a_{y}=-\mu v_{y}-g={\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}$ (2) Solving (1) is an elementarydifferential equation, thus the steps leading to a unique solution for`v_x` and, subsequently,`x` will not be enumerated. Given the initial conditions $v_{x}=v_{x0}$ (where`v_x0` is understood to be the x component of the initial velocity) and $x=0$ for $t=0$ : $v_{x}=v_{x0}e^{-\mu t}$ (1a)

Derivation of horizontal position

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

This implies that:

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

and

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

Solving (1) is an elementarydifferential equation, thus the steps leading to a unique solution forv_x and, subsequently,x will not be enumerated. Given the initial conditions $v_{x}=v_{x0}$ (wherev_x0 is understood to be the x component of the initial velocity) and $x=0$ for $t=0$ :

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

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 $v_{y}=v_{y0}$ and $y=0$ when $t=0$ . ${\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}=-\mu v_{y}-g$ (2) ${\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}+\mu v_{y}=-g$ (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 anintegrating factor $e^{\int \mu \,\mathrm {d} t}$ . $e^{\mu t}({\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}+\mu v_{y})=e^{\mu t}(-g)$ (2c) $(e^{\mu t}v_{y})^{\prime }=e^{\mu t}(-g)$ (2d) $\int {(e^{\mu t}v_{y})^{\prime }\,\mathrm {d} t}=e^{\mu t}v_{y}=\int {e^{\mu t}(-g)\,\mathrm {d} t}$ (2e) $e^{\mu t}v_{y}={\frac {1}{\mu }}e^{\mu t}(-g)+C$ (2f) $v_{y}={\frac {-g}{\mu }}+Ce^{-\mu t}$ (2g) And by integration we find: $y=-{\frac {g}{\mu }}t-{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}+C$ (3) Solving for our initial conditions: $v_{y}(t)=-{\frac {g}{\mu }}+(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}$ (2h) $y(t)=-{\frac {g}{\mu }}t-{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}+{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})$ (3a) With a bit of algebra to simplify (3a):

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 $v_{y}=v_{y0}$ and $y=0$ when $t=0$ .

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

${\frac {\mathrm {d} v_{y}}{\mathrm {d} t}}+\mu v_{y}=-g$ (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 anintegrating factor $e^{\int \mu \,\mathrm {d} t}$ .

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

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

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

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

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

And by integration we find:

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

Solving for our initial conditions:

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

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

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

y(t)=-{\frac {g}{\mu }}t+{\frac {1}{\mu }}\left(v_{y0}+{\frac {g}{\mu }}\right)(1-e^{-\mu t})

(3b)

Derivation of the time of flight
The total time of the journey in the presence of air resistance (more specifically, when $F_{air}=-kv$ ) can be calculated by the same strategy as above, namely, we solve the equation $y(t)=0$ . While in the case of zero air resistance this equation can be solved elementarily, here we shall need theLambert W function. The equation $y(t)=-{\frac {g}{\mu }}t+{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})(1-e^{-\mu t})=0$ is of the form $c_{1}t+c_{2}+c_{3}e^{c_{4}t}=0$ , and such an equation can be transformed into an equation solvable by the $W {\displaystyle W}$ function (see an example of such a transformationhere). Some algebra shows that the total time of flight, in closed form, is given as^[10]

Derivation of the time of flight

The total time of the journey in the presence of air resistance (more specifically, when $F_{air}=-kv$ ) can be calculated by the same strategy as above, namely, we solve the equation $y(t)=0$ . While in the case of zero air resistance this equation can be solved elementarily, here we shall need theLambert W function. The equation $y(t)=-{\frac {g}{\mu }}t+{\frac {1}{\mu }}(v_{y0}+{\frac {g}{\mu }})(1-e^{-\mu t})=0$ is of the form $c_{1}t+c_{2}+c_{3}e^{c_{4}t}=0$ , and such an equation can be transformed into an equation solvable by the $W {\displaystyle W}$ function (see an example of such a transformationhere). Some algebra shows that the total time of flight, in closed form, is given as^[10]

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

Trajectory of a projectile with Newton drag

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Trajectories of askydiver in air with Newton drag: para-flight; bomber near(ly above) target. ${\textstyle v.d=10^{3}.\nu =150}$ cm²/s?

Trajectories of askydiver in air with Newton drag: para-flight; bomber near(ly above) target. ${\textstyle v.d=10^{3}.\nu =150}$ cm²/s?

The most typical case ofair resistance, in case ofReynolds numbers above about 1000, is Newton drag with a drag force proportional to the speed squared, $F_{\mathrm {air} }=-kv^{2}$ . In air, which has akinematic viscosity around 0.15 cm²/s, this means that the product of object speed and diameter must be more than about 0.015 m²/s.

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

The following assumptions are made:

Constantgravitational acceleration ${\textstyle g}$
Air resistance is given by the followingdrag formula,

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

Where:

F_D is the drag force,
c is thedrag coefficient,
ρ is theair density,
A is thecross sectional area of the projectile. Again $\mu =k/m:=$ $c\rho A/(2m)$ $=c\rho /(2\rho _{p}l)$ . Compare this with theory/practice of theballistic coefficient.

Special cases

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Even though the general case of a projectile with Newton drag cannot be solved analytically, some special cases can. Here we denote theterminal velocity in free-fall as ${\textstyle v_{\infty }={\sqrt {g/\mu }}}$ and the characteristicsettling time constant ${\textstyle t_{f}\equiv 1/(v_{\infty }\mu )=1/{\sqrt {g\mu }}}$ . (Dimension of $g {\displaystyle g}$ [m/s²], $\mu$ [1/m])

Near-horizontal motion: In case the motion is almost horizontal, $|v_{x}|\gg |v_{y}|$ , such as a flying bullet. The vertical velocity component has very little influence on the horizontal motion. In this case:^[11]

{\dot {v}}_{x}(t)=-\mu \,v_{x}^{2}(t)

v_{x}(t)={\frac {1}{1/v_{x,0}+\mu \,t}}

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 (relatively small

g {\displaystyle g}

). It also applies when vertical motion is prevented, such as for a moving car with its engine off.

Vertical motionupward:^[11]

{\dot {v}}_{y}(t)=-g-\mu \,v_{y}^{2}(t)

v_{y}(t)=v_{\infty }\tan {\frac {t_{\mathrm {peak} }-t}{t_{f}}}

y(t)=y_{\mathrm {peak} }+{\frac {1}{\mu }}\ln {\bigl (}\cos {\frac {t_{\mathrm {peak} }-t}{t_{f}}}{\bigr )}

Here

v_{\infty }\equiv {\sqrt {\frac {g}{\mu }}}

and

t_{f}={\frac {1}{\sqrt {\mu g}}},

t_{\mathrm {peak} }\equiv t_{f}\arctan {\frac {v_{y,0}}{v_{\infty }}}={\frac {1}{\sqrt {\mu g}}}\arctan {\left({\sqrt {\frac {\mu }{g}}}v_{y,0}\right)},

and

y_{\mathrm {peak} }\equiv -{\frac {1}{\mu }}\ln {\cos {\frac {t_{\mathrm {peak} }}{t_{f}}}}={\frac {1}{2\mu }}\ln {{\bigl (}1+{\frac {\mu }{g}}v_{y,0}^{2}{\bigr )}}

where

v_{y,0}

is the initial upward velocity at

t=0

and the initial position is

y(0)=0

A projectile cannot rise longer than

t_{\mathrm {rise} }={\frac {\pi }{2}}t_{f}

in the vertical direction, when it reaches the peak (0 m, y_peak) at 0 m/s.

Vertical motiondownward:^[11]

{\dot {v}}_{y}(t)=-g+\mu \,v_{y}^{2}(t)

v_{y}(t)=-v_{\infty }\tanh {\frac {t-t_{\mathrm {peak} }}{t_{f}}}

y(t)=y_{\mathrm {peak} }-{\frac {1}{\mu }}\ln(\cosh {\biggl (}{\frac {t-t_{\mathrm {peak} }}{t_{f}}}{\biggr )})

Withhyperbolic functions

After a time

t_{f}

at y=0, the projectile reaches almost terminal velocity

-v_{\infty }

Numerical solution

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

{\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 (in product $c(v)\rho (y)$ ) and position-dependent gravity field ${\textstyle g(y)=g_{0}/(1+y/R)^{2}}$ (when ${\textstyle y\ll R:g\lesssim g_{0}/(1+2y/R)\approx g_{0}(1-2y/R)}$ , is linear decrease).

Lofted trajectory

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Lofted trajectories of North Korean ballistic missilesHwasong-14,Hwasong-15 andHwasong-17

A special case of a ballistic trajectory for a rocket is a lofted trajectory, a trajectory with anapogee greater than theminimum-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 inspaceflight.^[12]

Projectile motion on a planetary scale

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

When a projectile travels a range that is significant compared to the Earth's radius (above ≈100 km), thecurvature of the Earth and the non-uniformEarth's gravity have to be considered. This is, for example, the case with spacecrafts and intercontinentalmissiles. The trajectory then generalizes (without air resistance) from a parabola to a Kepler-ellipse with one focus at the center of the Earth (shown in fig. 3). The projectile motion then followsKepler's laws of planetary motion.

The trajectory's parameters have to be adapted from the values of a uniform gravity field stated above. TheEarth radius is taken asR, andg as the standard surface gravity. Let ${\tilde {v}}:=v/{\sqrt {Rg}}$ be the launch velocity relative to the first cosmic orescape velocity.