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Projectile motion is a form of motion experienced by an object or particle (a projectile) that is projected near 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 straight 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.

Parabolic water motion trajectory
Components of initial velocity of parabolic throwing

Ballistics (from Ancient Greek βάλλεινbállein '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.

Contents

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, and used by him to prove the parabolic form of projectile motion.

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 v 0 cos θ {\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. The components of the acceleration are:

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

Velocity

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

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

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

v 0 x = v 0 cos ( θ ) {\displaystyle v_{0x}=v_{0}\cos(\theta )} ,
v 0 y = v 0 sin ( θ ) {\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, 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:

v x = v 0 cos ( θ ) {\displaystyle v_{x}=v_{0}\cos(\theta )} ,
v y = v 0 sin ( θ ) g t {\displaystyle v_{y}=v_{0}\sin(\theta )-gt} .

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

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

Displacement

Displacement and coordinates of parabolic throwing

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

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

The magnitude of the displacement is:

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

Consider the equations,

x = v 0 t cos ( θ ) , y = v 0 t sin ( θ ) 1 2 g t 2 {\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:

y = tan ( θ ) x g 2 v 0 2 cos 2 θ x 2 {\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

y = a x + b x 2 {\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:

v 0 = x 2 g x sin 2 θ 2 y cos 2 θ {\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

r ( ϕ ) = v 0 2 cos 2 θ g ( tan ϕ sec ϕ + ( tan 2 ϕ + tan 2 θ ) sec 2 ϕ ) {\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 0 ϕ π {\displaystyle 0\leq \phi \leq \pi } . This expression can be obtained by transforming the Cartesian equation as stated above by y = r sin ϕ {\displaystyle y=r\sin \phi } and x = r cos ϕ {\displaystyle x=r\cos \phi } .

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.

y = v 0 t sin ( θ ) 1 2 g t 2 {\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 y = 0 {\displaystyle y=0} .

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

t = d v cos θ = v sin θ + ( v sin θ ) 2 + 2 g y 0 g {\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

t = v sin θ + ( v sin θ ) 2 g = v sin θ + v sin θ g = 2 v sin θ g = 2 v sin ( 45 ) g = 2 v 2 2 g = 2 v g {\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.

Time of flight to the target's position

As shown above in the Displacement 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 ( θ ) {\displaystyle x=v_{0}t\cos(\theta )}


x t = v 0 cos ( θ ) {\displaystyle {\frac {x}{t}}=v_{0}\cos(\theta )}


t = x v 0 cos ( θ ) {\displaystyle t={\frac {x}{v_{0}\cos(\theta )}}}

This equation will give the total time t the projectile must travel for to reach the target's horizontal displacement, neglecting air resistance.

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 v y = 0 {\displaystyle v_{y}=0} , that is,

0 = v 0 sin ( θ ) g t h {\displaystyle 0=v_{0}\sin(\theta )-gt_{h}} .

Time to reach the maximum height(h):

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

For the vertical displacement of the maximum height of projectile:

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

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

h m a x = v 0 2 2 g {\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 t d 2 {\displaystyle {\frac {t_{d}}{2}}} is:

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

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

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

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

Maximum distance of projectile

Main article: Range of a 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 ( y = 0 {\displaystyle y=0} ).

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

Time to reach ground:

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

From the horizontal displacement the maximum distance of projectile:

d = v 0 t d cos ( θ ) {\displaystyle d=v_{0}t_{d}\cos(\theta )} ,

so

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

Note that d has its maximum value when

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

which necessarily corresponds to

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

or

θ = 45 {\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.

d = v cos θ g ( v sin θ + ( v sin θ ) 2 + 2 g y 0 ) {\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:

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

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

d m a x = v 2 g {\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:

v y 2 = ( v 0 sin θ ) 2 2 g y {\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.

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

There are two solutions:

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

and

θ = 1 2 arccos ( g d v 2 ) {\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:

θ = arctan ( v 2 ± v 4 g ( g x 2 + 2 y v 2 ) g x ) {\displaystyle \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 y = 0 {\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 v 2 {\displaystyle v^{2}} , and we find

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

This gives

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

If we denote the angle whose tangent isy/x byα, then

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

This implies

θ = π / 2 1 2 ( π / 2 α ) . {\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:

L = v 0 2 2 g ( 2 sin θ + cos 2 θ ln 1 + sin θ 1 sin θ ) = v 0 2 g ( sin θ + cos 2 θ tanh 1 ( sin θ ) ) {\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 v 0 {\displaystyle v_{0}} is the initial velocity, θ {\displaystyle \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 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:

L = 0 r a n g e 1 + ( d y d x ) 2 d x = 0 v 0 2 sin ( 2 θ ) / g 1 + ( g v 0 2 cos 2 θ x + tan θ ) 2 d x {\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} .
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: F a i r = f ( v ) v ^ {\displaystyle \mathbf {F_{air}} =-f(v)\cdot \mathbf {\hat {v}} } . The speed-dependence of the friction force is linear ( f ( v ) v {\displaystyle f(v)\propto v} ) at very low speeds (Stokes drag) and quadratic ( f ( v ) v 2 {\displaystyle f(v)\propto v^{2}} ) at larger speeds (Newton drag). 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 0.15 c m 2 / s {\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 0.015 m 2 / s {\displaystyle 0.015\,\mathrm {m^{2}/s} } , which is typically the case for projectiles.

  • Stokes drag: F a i r = k S t o k e s v {\displaystyle \mathbf {F_{air}} =-k_{\mathrm {Stokes} }\cdot \mathbf {v} \qquad } (for R e 1000 {\displaystyle Re\lesssim 1000} )
  • Newton drag: F a i r = k | v | v {\displaystyle \mathbf {F_{air}} =-k\,|\mathbf {v} |\cdot \mathbf {v} \qquad } (for R e 1000 {\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: F a i r = f ( v ) v ^ {\displaystyle \mathbf {F_{\mathrm {air} }} =-f(v)\cdot \mathbf {\hat {v}} }

Trajectory of a projectile with Stokes drag

Stokes drag, where F a i r v {\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 F a i r {\displaystyle F_{\mathrm {air} }} on v {\displaystyle v} causes a very simple differential equation of motion

d d t ( v x v y ) = ( μ v x g μ v y ) {\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. Here, v 0 {\displaystyle v_{0}} , v x {\displaystyle v_{x}} and v y {\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 μ := k / m {\displaystyle \mu :=k/m} . For the derivation only the case where 0 o θ 180 o {\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 Σ F = k v x = m a x {\displaystyle \Sigma F=-kv_{x}=ma_{x}} and in the y direction Σ F = k v y m g = m a y {\displaystyle \Sigma F=-kv_{y}-mg=ma_{y}} .

This implies that:

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

and

a y = μ v y g = d v y d t {\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 v x = v x 0 {\displaystyle v_{x}=v_{x0}} (where vx0 is understood to be the x component of the initial velocity) and x = 0 {\displaystyle x=0} for t = 0 {\displaystyle t=0} :

v x = v x 0 e μ t {\displaystyle v_{x}=v_{x0}e^{-\mu t}} (1a)

x ( t ) = v x 0 μ ( 1 e μ t ) {\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 v y = v y 0 {\displaystyle v_{y}=v_{y0}} and y = 0 {\displaystyle y=0} when t = 0 {\displaystyle t=0} .

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

d v y d t + μ v y = g {\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 e μ d t {\displaystyle e^{\int \mu \,\mathrm {d} t}} .

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

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

( e μ t v y ) d t = e μ t v y = e μ t ( g ) d t {\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)

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

v y = g μ + C e μ t {\displaystyle v_{y}={\frac {-g}{\mu }}+Ce^{-\mu t}} (2g)

And by integration we find:

y = g μ t 1 μ ( v y 0 + g μ ) e μ t + C {\displaystyle 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 ) = g μ + ( v y 0 + g μ ) e μ t {\displaystyle v_{y}(t)=-{\frac {g}{\mu }}+(v_{y0}+{\frac {g}{\mu }})e^{-\mu t}} (2h)

y ( t ) = g μ t 1 μ ( v y 0 + g μ ) e μ t + 1 μ ( v y 0 + g μ ) {\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):

y ( t ) = g μ t + 1 μ ( v y 0 + g μ ) ( 1 e μ t ) {\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 F a i r = k v {\displaystyle F_{air}=-kv} ) can be calculated by the same strategy as above, namely, we solve the equation y ( t ) = 0 {\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 y ( t ) = g μ t + 1 μ ( v y 0 + g μ ) ( 1 e μ t ) = 0 {\displaystyle 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 {\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 W {\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

t = 1 μ ( 1 + μ g v y 0 + W ( ( 1 + μ g v y 0 ) e ( 1 + μ g v y 0 ) ) ) {\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, F a i r = k v 2 {\displaystyle F_{\mathrm {air} }=-kv^{2}} . In air, which has a kinematic viscosity around 0.15 c m 2 / s {\displaystyle 0.15\,\mathrm {cm^{2}/s} } , this means that the product of speed and diameter must be more than about 0.015 m 2 / s {\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.

The following assumptions are made:

F D = 1 2 c ρ A v v {\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 v = g / μ {\displaystyle v_{\infty }={\sqrt {g/\mu }}} and the characteristic settling time constant t f = 1 / g μ {\displaystyle t_{f}=1/{\sqrt {g\mu }}} .

  • Near-horizontal motion: In case the motion is almost horizontal, | v x | | v y | {\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:
v ˙ x ( t ) = μ v x 2 ( t ) {\displaystyle {\dot {v}}_{x}(t)=-\mu \,v_{x}^{2}(t)}
v x ( t ) = 1 1 / v x , 0 + μ t {\displaystyle v_{x}(t)={\frac {1}{1/v_{x,0}+\mu \,t}}}
x ( t ) = 1 μ ln ( 1 + μ v x , 0 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:
v ˙ y ( t ) = g μ v y 2 ( t ) {\displaystyle {\dot {v}}_{y}(t)=-g-\mu \,v_{y}^{2}(t)}
v y ( t ) = v tan t p e a k t t f {\displaystyle v_{y}(t)=v_{\infty }\tan {\frac {t_{\mathrm {peak} }-t}{t_{f}}}}
y ( t ) = y p e a k + 1 μ ln ( cos t p e a k 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 t r i s e = π 2 t f {\displaystyle t_{\mathrm {rise} }={\frac {\pi }{2}}t_{f}} vertically before it reaches the peak.
  • Vertical motion downward:
v ˙ y ( t ) = g + μ v y 2 ( t ) {\displaystyle {\dot {v}}_{y}(t)=-g+\mu \,v_{y}^{2}(t)}
v y ( t ) = v tanh t t p e a k t f {\displaystyle v_{y}(t)=-v_{\infty }\tanh {\frac {t-t_{\mathrm {peak} }}{t_{f}}}}
y ( t ) = y p e a k 1 μ ln ( cosh t t p e a k 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 t f {\displaystyle t_{f}} , the projectile reaches almost terminal velocity v {\displaystyle -v_{\infty }} .


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

d d t ( x y v x v y ) = ( v x v y μ v x v x 2 + v y 2 g μ v y v x 2 + v y 2 ) {\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 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.

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 Earth's gravity 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 v ~ := v / R g {\displaystyle {\tilde {v}}:=v/{\sqrt {Rg}}} the launch velocity relative to the first cosmic velocity.

Total range d between launch and impact:

d = v 2 sin ( 2 θ ) g / 1 ( 2 v ~ 2 ) v ~ 2 cos 2 θ {\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 ( θ = 1 2 arccos ( v ~ 2 / ( 2 v ~ 2 ) ) {\displaystyle \theta ={\tfrac {1}{2}}\arccos \left({\tilde {v}}^{2}/(2-{\tilde {v}}^{2})\right)} ):

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

Maximum height of a projectile above the planetary surface:

h = v 2 sin 2 θ g / ( 1 v ~ 2 + 1 ( 2 v ~ 2 ) v ~ 2 cos 2 θ ) {\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 ( θ = 90 {\displaystyle \theta =90^{\circ }} ):

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

Time of flight:

t = 2 v sin θ g 1 2 v ~ 2 ( 1 + 1 2 v ~ 2 v ~ sin θ arcsin 2 v ~ 2 v ~ sin θ 1 ( 2 v ~ 2 ) v ~ 2 cos 2 θ ) {\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)}
  1. The g is the acceleration due to gravity. ( 9.81 m / s 2 {\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. 2 sin ( α ) cos ( α ) = sin ( 2 α ) {\displaystyle 2\cdot \sin(\alpha )\cdot \cos(\alpha )=\sin(2\alpha )}
  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. Walter Greiner (2004). Classical Mechanics: Point Particles and Relativity. Springer Science & Business Media. p. 181. ISBN 0-387-95586-0.
  8. Ballistic Missile Defense, Glossary, v. 3.0, US Department of Defense, June 1997.

Projectile motion Article Talk Language Watch Edit 160 160 Redirected from Trajectory of a projectile This article has multiple issues Please help improve it or discuss these issues on the talk page Learn how and when to remove these template messages This article needs attention from an expert in Physics The specific problem is Contains various high level info without any references WikiProject Physics may be able to help recruit an expert November 2019 This article needs additional citations for verification Please help improve this article by adding citations to reliable sources Unsourced material may be challenged and removed Find sources Projectile motion news newspapers books scholar JSTOR November 2019 Learn how and when to remove this template message Learn how and when to remove this template message Projectile motion is a form of motion experienced by an object or particle a projectile that is projected near 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 straight 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 Parabolic water motion trajectory Components of initial velocity of parabolic throwing Ballistics from Ancient Greek ballein ballein 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 Contents 1 Kinematic quantities of projectile motion 1 1 Acceleration 1 2 Velocity 1 3 Displacement 1 4 Displacement in polar coordinates 2 Properties of the trajectory 2 1 Time of flight or total time of the whole journey 2 2 Time of flight to the target s position 2 3 Maximum height of projectile 2 4 Relation between horizontal range and maximum height 2 5 Maximum distance of projectile 2 6 Application of the work energy theorem 2 7 Angle of reach 2 8 Angle 8 required to hit coordinate x y 2 9 Total Path Length of the Trajectory 3 Trajectory of a projectile with air resistance 3 1 Trajectory of a projectile with Stokes drag 3 2 Trajectory of a projectile with Newton drag 3 2 1 Special cases 3 2 2 Numerical solution 4 Lofted trajectory 5 Projectile motion on a planetary scale 6 Notes 7 ReferencesKinematic quantities of projectile motion EditIn 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 Edit Since there is only acceleration in the vertical direction the velocity in the horizontal direction is constant being equal to v 0 cos 8 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 a x 0 displaystyle a x 0 a y g displaystyle a y g Velocity Edit Let the projectile be launched with an initial velocity v 0 v 0 displaystyle mathbf v 0 equiv mathbf v 0 which can be expressed as the sum of horizontal and vertical components as follows v 0 v 0 x x v 0 y y displaystyle mathbf v 0 v 0x mathbf hat x v 0y mathbf hat y The components v 0 x displaystyle v 0x and v 0 y displaystyle v 0y can be found if the initial launch angle 8 displaystyle theta is known v 0 x v 0 cos 8 displaystyle v 0x v 0 cos theta v 0 y v 0 sin 8 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 v x v 0 cos 8 displaystyle v x v 0 cos theta v y v 0 sin 8 g t displaystyle v y v 0 sin theta gt The magnitude of the velocity under the Pythagorean theorem also known as the triangle law v v x 2 v y 2 displaystyle v sqrt v x 2 v y 2 Displacement Edit Displacement and coordinates of parabolic throwing At any time t displaystyle t the projectile s horizontal and vertical displacement are x v 0 t cos 8 displaystyle x v 0 t cos theta y v 0 t sin 8 1 2 g t 2 displaystyle y v 0 t sin theta frac 1 2 gt 2 The magnitude of the displacement is D r x 2 y 2 displaystyle Delta r sqrt x 2 y 2 Consider the equations x v 0 t cos 8 y v 0 t sin 8 1 2 g t 2 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 y tan 8 x g 2 v 0 2 cos 2 8 x 2 displaystyle y tan theta cdot x frac g 2v 0 2 cos 2 theta cdot x 2 Since g 8 and v0 are constants the above equation is of the form y a x b x 2 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 8 or a are known the initial velocity can be found solving for v0 in the aforementioned parabolic equation v 0 x 2 g x sin 2 8 2 y cos 2 8 displaystyle v 0 sqrt x 2 g over x sin 2 theta 2y cos 2 theta Displacement in polar coordinates Edit 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 r ϕ v 0 2 cos 2 8 g tan ϕ sec ϕ tan 2 ϕ tan 2 8 sec 2 ϕ 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 0 ϕ p displaystyle 0 leq phi leq pi This expression can be obtained by transforming the Cartesian equation as stated above by y r sin ϕ displaystyle y r sin phi and x r cos ϕ displaystyle x r cos phi Properties of the trajectory EditTime of flight or total time of the whole journey Edit The total time t for which the projectile remains in the air is called the time of flight y v 0 t sin 8 1 2 g t 2 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 y 0 displaystyle y 0 0 v 0 t sin 8 1 2 g t 2 displaystyle 0 v 0 t sin theta frac 1 2 gt 2 v 0 t sin 8 1 2 g t 2 displaystyle v 0 t sin theta frac 1 2 gt 2 v 0 sin 8 1 2 g t displaystyle v 0 sin theta frac 1 2 gt t 2 v 0 sin 8 g 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 t d v cos 8 v sin 8 v sin 8 2 2 g y 0 g 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 t v sin 8 v sin 8 2 g v sin 8 v sin 8 g 2 v sin 8 g 2 v sin 45 g 2 v 2 2 g 2 v g 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 8 is 45 and y0 is 0 Time of flight to the target s position Edit As shown above in the Displacement 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 8 displaystyle x v 0 t cos theta x t v 0 cos 8 displaystyle frac x t v 0 cos theta t x v 0 cos 8 displaystyle t frac x v 0 cos theta This equation will give the total time t the projectile must travel for to reach the target s horizontal displacement neglecting air resistance Maximum height of projectile Edit 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 v y 0 displaystyle v y 0 that is 0 v 0 sin 8 g t h displaystyle 0 v 0 sin theta gt h Time to reach the maximum height h t h v 0 sin 8 g displaystyle t h frac v 0 sin theta g For the vertical displacement of the maximum height of projectile h v 0 t h sin 8 1 2 g t h 2 displaystyle h v 0 t h sin theta frac 1 2 gt h 2 h v 0 2 sin 2 8 2 g displaystyle h frac v 0 2 sin 2 theta 2g The maximum reachable height is obtained for 8 90 h m a x v 0 2 2 g displaystyle h mathrm max frac v 0 2 2g Relation between horizontal range and maximum height Edit The relation between the range d on the horizontal plane and the maximum height h reached at t d 2 displaystyle frac t d 2 is h d tan 8 4 displaystyle h frac d tan theta 4 Proofh v 0 2 sin 2 8 2 g displaystyle h frac v 0 2 sin 2 theta 2g d v 0 2 sin 2 8 g displaystyle d frac v 0 2 sin 2 theta g h d v 0 2 sin 2 8 2 g displaystyle frac h d frac v 0 2 sin 2 theta 2g g v 0 2 sin 2 8 displaystyle frac g v 0 2 sin 2 theta h d sin 2 8 4 sin 8 cos 8 displaystyle frac h d frac sin 2 theta 4 sin theta cos theta h d tan 8 4 displaystyle h frac d tan theta 4 Maximum distance of projectile Edit Main article Range of a 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 y 0 displaystyle y 0 0 v 0 t d sin 8 1 2 g t d 2 displaystyle 0 v 0 t d sin theta frac 1 2 gt d 2 Time to reach ground t d 2 v 0 sin 8 g displaystyle t d frac 2v 0 sin theta g From the horizontal displacement the maximum distance of projectile d v 0 t d cos 8 displaystyle d v 0 t d cos theta so note 3 d v 0 2 g sin 2 8 displaystyle d frac v 0 2 g sin 2 theta Note that d has its maximum value when sin 2 8 1 displaystyle sin 2 theta 1 which necessarily corresponds to 2 8 90 displaystyle 2 theta 90 circ or 8 45 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 d v cos 8 g v sin 8 v sin 8 2 2 g y 0 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 d v 2 sin 2 8 g displaystyle d frac v 2 sin 2 theta g Thus the maximum distance is obtained if 8 is 45 degrees This distance is d m a x v 2 g displaystyle d mathrm max frac v 2 g Application of the work energy theorem Edit According to the work energy theorem the vertical component of velocity is v y 2 v 0 sin 8 2 2 g y 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 Edit The angle of reach is the angle 8 at which a projectile must be launched in order to go a distance d given the initial velocity v sin 2 8 g d v 2 displaystyle sin 2 theta frac gd v 2 There are two solutions 8 1 2 arcsin g d v 2 displaystyle theta frac 1 2 arcsin left frac gd v 2 right shallow trajectory and 8 1 2 arccos g d v 2 displaystyle theta frac 1 2 arccos left frac gd v 2 right steep trajectory Angle 8 required to hit coordinate x y Edit 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 8 are 8 arctan v 2 v 4 g g x 2 2 y v 2 g x displaystyle 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 y 0 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 v 2 displaystyle v 2 and we find v 2 g y y 2 x 2 displaystyle v 2 g y sqrt y 2 x 2 This gives 8 arctan y x y 2 x 2 1 displaystyle theta arctan left y x sqrt y 2 x 2 1 right If we denote the angle whose tangent is y x by a then tan 8 sin a 1 cos a displaystyle tan theta frac sin alpha 1 cos alpha tan p 2 8 cos a sin a 1 displaystyle tan pi 2 theta frac cos alpha sin alpha 1 cos 2 p 2 8 1 2 sin a 1 displaystyle cos 2 pi 2 theta frac 1 2 sin alpha 1 2 cos 2 p 2 8 1 cos p 2 a displaystyle 2 cos 2 pi 2 theta 1 cos pi 2 alpha This implies 8 p 2 1 2 p 2 a 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 Edit 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 L v 0 2 2 g 2 sin 8 cos 2 8 ln 1 sin 8 1 sin 8 v 0 2 g sin 8 cos 2 8 tanh 1 sin 8 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 v 0 displaystyle v 0 is the initial velocity 8 displaystyle 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 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 L 0 r a n g e 1 d y d x 2 d x 0 v 0 2 sin 2 8 g 1 g v 0 2 cos 2 8 x tan 8 2 d x 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 Edit 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 F a i r f v v displaystyle mathbf F air f v cdot mathbf hat v The speed dependence of the friction force is linear f v v displaystyle f v propto v at very low speeds Stokes drag and quadratic f v v 2 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 0 15 c m 2 s 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 0 015 m 2 s displaystyle 0 015 mathrm m 2 s which is typically the case for projectiles Stokes drag F a i r k S t o k e s v displaystyle mathbf F air k mathrm Stokes cdot mathbf v qquad for R e 1000 displaystyle Re lesssim 1000 Newton drag F a i r k v v displaystyle mathbf F air k mathbf v cdot mathbf v qquad for R e 1000 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 F a i r f v v displaystyle mathbf F mathrm air f v cdot mathbf hat v Trajectory of a projectile with Stokes drag Edit Stokes drag where F a i r v 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 F a i r displaystyle F mathrm air on v displaystyle v causes a very simple differential equation of motion d d t v x v y m v x g m v y 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 v 0 displaystyle v 0 v x displaystyle v x and v y 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 m k m displaystyle mu k m For the derivation only the case where 0 o 8 180 o displaystyle 0 o leq theta leq 180 o is considered Again the projectile is fired from the origin 0 0 Derivation of horizontal positionThe 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 S F k v x m a x displaystyle Sigma F kv x ma x and in the y direction S F k v y m g m a y displaystyle Sigma F kv y mg ma y This implies that a x m v x d v x d t displaystyle a x mu v x frac mathrm d v x mathrm d t 1 and a y m v y g d v y d t 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 v x v x 0 displaystyle v x v x0 where vx0 is understood to be the x component of the initial velocity and x 0 displaystyle x 0 for t 0 displaystyle t 0 v x v x 0 e m t displaystyle v x v x0 e mu t 1a x t v x 0 m 1 e m t displaystyle x t frac v x0 mu left 1 e mu t right 1b Derivation of vertical positionWhile 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 y 0 displaystyle v y v y0 and y 0 displaystyle y 0 when t 0 displaystyle t 0 d v y d t m v y g displaystyle frac mathrm d v y mathrm d t mu v y g 2 d v y d t m v y g 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 e m d t displaystyle e int mu mathrm d t e m t d v y d t m v y e m t g displaystyle e mu t frac mathrm d v y mathrm d t mu v y e mu t g 2c e m t v y e m t g displaystyle e mu t v y prime e mu t g 2d e m t v y d t e m t v y e m t g d t 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 e m t v y 1 m e m t g C displaystyle e mu t v y frac 1 mu e mu t g C 2f v y g m C e m t displaystyle v y frac g mu Ce mu t 2g And by integration we find y g m t 1 m v y 0 g m e m t C displaystyle 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 g m v y 0 g m e m t displaystyle v y t frac g mu v y0 frac g mu e mu t 2h y t g m t 1 m v y 0 g m e m t 1 m v y 0 g m 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 y t g m t 1 m v y 0 g m 1 e m t 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 flightThe total time of the journey in the presence of air resistance more specifically when F a i r k v displaystyle F air kv can be calculated by the same strategy as above namely we solve the equation y t 0 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 y t g m t 1 m v y 0 g m 1 e m t 0 displaystyle 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 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 W 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 t 1 m 1 m g v y 0 W 1 m g v y 0 e 1 m g v y 0 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 Edit 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 F a i r k v 2 displaystyle F mathrm air kv 2 In air which has a kinematic viscosity around 0 15 c m 2 s displaystyle 0 15 mathrm cm 2 s this means that the product of speed and diameter must be more than about 0 015 m 2 s 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 The following assumptions are made Constant gravitational acceleration Air resistance is given by the following drag formula F D 1 2 c r A v v displaystyle mathbf F D tfrac 1 2 c rho A v mathbf v dd Where dd FD is the drag force c is the drag coefficient r is the air density A is the cross sectional area of the projectile m k m crA 2m dd Special cases Edit 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 v g m displaystyle v infty sqrt g mu and the characteristic settling time constant t f 1 g m displaystyle t f 1 sqrt g mu Near horizontal motion In case the motion is almost horizontal v x v y 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 v x t m v x 2 t displaystyle dot v x t mu v x 2 t v x t 1 1 v x 0 m t displaystyle v x t frac 1 1 v x 0 mu t x t 1 m ln 1 m v x 0 t displaystyle x t frac 1 mu ln 1 mu v x 0 cdot t dd 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 v y t g m v y 2 t displaystyle dot v y t g mu v y 2 t v y t v tan t p e a k t t f displaystyle v y t v infty tan frac t mathrm peak t t f y t y p e a k 1 m ln cos t p e a k t t f displaystyle y t y mathrm peak frac 1 mu ln left cos frac t mathrm peak t t f right dd A projectile can not rise longer than t r i s e p 2 t f displaystyle t mathrm rise frac pi 2 t f vertically before it reaches the peak Vertical motion downward 7 v y t g m v y 2 t displaystyle dot v y t g mu v y 2 t v y t v tanh t t p e a k t f displaystyle v y t v infty tanh frac t t mathrm peak t f y t y p e a k 1 m ln cosh t t p e a k t f displaystyle y t y mathrm peak frac 1 mu ln left cosh frac t t mathrm peak t f right dd After a time t f displaystyle t f the projectile reaches almost terminal velocity v displaystyle v infty Numerical solution Edit 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 d d t x y v x v y v x v y m v x v x 2 v y 2 g m v y v x 2 v y 2 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 Edit 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 Edit 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 Earth s gravity 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 v v R g displaystyle tilde v v sqrt Rg the launch velocity relative to the first cosmic velocity Total range d between launch and impact d v 2 sin 2 8 g 1 2 v 2 v 2 cos 2 8 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 8 1 2 arccos v 2 2 v 2 displaystyle theta tfrac 1 2 arccos left tilde v 2 2 tilde v 2 right d m a x v 2 g 1 1 2 v 2 displaystyle d mathrm max frac v 2 g big left 1 tfrac 1 2 tilde v 2 right with v lt R g displaystyle v lt sqrt Rg the first cosmic velocity Maximum height of a projectile above the planetary surface h v 2 sin 2 8 g 1 v 2 1 2 v 2 v 2 cos 2 8 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 8 90 displaystyle theta 90 circ h m a x v 2 2 g 1 1 2 v 2 displaystyle h mathrm max frac v 2 2g big left 1 tfrac 1 2 tilde v 2 right with v lt 2 R g displaystyle v lt sqrt 2Rg the second cosmic velocity Time of flight t 2 v sin 8 g 1 2 v 2 1 1 2 v 2 v sin 8 arcsin 2 v 2 v sin 8 1 2 v 2 v 2 cos 2 8 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 Edit The g is the acceleration due to gravity 9 81 m s 2 displaystyle 9 81 mathrm m s 2 near the surface of the Earth decreasing when the object goes upward and increasing when it goes downward 2 sin a cos a sin 2 a displaystyle 2 cdot sin alpha cdot cos alpha sin 2 alpha References Edit Galileo Galilei Two New Sciences Leiden 1638 p 249 Nolte David D Galileo Unbound Oxford University Press 2018 pp 39 63 Tatum 2019 Classical Mechanics PDF pp ch 7 Stephen T Thornton Jerry B Marion 2007 Classical Dynamics of Particles and Systems Brooks Cole p 59 ISBN 978 0 495 55610 7 Atam P Arya Atam Parkash Arya September 1997 Introduction to Classical Mechanics Prentice Hall Internat p 227 ISBN 978 0 13 906686 3 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 a b c Walter Greiner 2004 Classical Mechanics Point Particles and Relativity Springer Science amp Business Media p 181 ISBN 0 387 95586 0 Ballistic Missile Defense Glossary v 3 0 US Department of Defense June 1997 Retrieved from https en wikipedia org w index php title Projectile motion amp oldid 1090776713, wikipedia, wiki, book,

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