# Trajectory of a projectile

Trajectories of a projectile with air drag and varying initial velocities

In physics, the ballistic trajectory of a projectile is the path that a thrown or launched projectile or missile will take under the action of gravity, neglecting all other forces, such as friction from aerodynamic drag, without propulsion.

The United States Department of Defense and NATO define a ballistic trajectory as a trajectory traced after the propulsive force is terminated and the body is acted upon only by gravity and aerodynamic drag.[1]

The following applies for ranges which are small compared to the size of the Earth. For longer ranges see sub-orbital spaceflight.

## Notation

In the equations on this page, the following variables will be used:

• g: the gravitational acceleration—usually taken to be 9.81 m/s2 near the Earth's surface
• θ: the angle at which the projectile is launched
• v: the speed at which the projectile is launched
• y0: the initial height of the projectile
• d: the total horizontal distance traveled by the projectile

Ballistics (gr. βάλλειν ('ba'llein'), "to throw") is the science of mechanics that deals with the flight, behavior, and effects of projectiles, especially bullets, gravity bombs, rockets, or the like; the science or art of designing and accelerating projectiles so as to achieve a desired performance. A ballistic body is a body which is free to move, behave, and be modified in appearance, contour, or texture by ambient conditions, substances, or forces, as by the pressure of gases in a gun, by rifling in a barrel, by gravity, by temperature, or by air particles. A ballistic missile is a missile only guided during the relatively brief initial powered phase of flight, whose course is subsequently governed by the laws of classical mechanics.

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

## Conditions at the final position of the projectile

### Distance traveled

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.

$d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)$

When the surface the object is launched from and is flying over is flat (the initial height is zero), the distance traveled is:

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

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

$d = \frac{v^2}{g}$

For explicit derivations of these results, see Range of a projectile.

### Time of flight

The time of flight (t) is the time it takes for the projectile to finish its trajectory.

$t = \frac{d}{v \cos\theta} = \frac{v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0}}{g}$

As above, this expression can be reduced to

$t = \frac{\sqrt{2} \cdot v}{g}$

if θ is 45° and y0 is 0.

The above results are found in Range of a projectile.

### 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.

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

## Conditions at an arbitrary distance x

### Height at x

The height y of the projectile at distance x is given by

$y = y_0 + x \tan \theta - \frac {gx^2}{2(v\cos\theta)^2}$.

The third term is the deviation from traveling in a straight line.

### Velocity at x

The magnitude, $|v|,$ of the velocity of the projectile at distance x is given by

$| v | = \sqrt{v^2 - 2gx \tan \theta + \left(\frac{gx}{v\cos \theta}\right)^2}$.

#### Derivation

The magnitude |v| of the velocity is given by

$| v | = \sqrt{V_x^2 + V_y^2}$,

where Vx and Vy are the instantaneous velocities in the x- and y-directions, respectively.

Here the x-velocity remains constant; it is always equal to v cos θ.

The y-velocity can be found using the formula

$v_f = v_i + at$

by setting vi = v sin θ, a = -g, and $t = \frac{x}{v \cos \theta}$. (The latter is found by taking x = (v cos θ) t and solving for t.) Then,

$V_y = v \sin \theta - \frac{gx}{v \cos \theta}$

and

$| v | = \sqrt{(v \cos \theta)^2 + \left(v \sin \theta - \frac{gx}{v \cos \theta} \right)^2}$.

The formula above is found by simplifying.

## Angle $\theta$ 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 $\theta$ 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 y = 0.

Derivation

First, two elementary formulae are called upon relating to projectile motion:

$x = v t \cos \theta$ (1)
$y = vt \sin \theta - \frac{1}{2} g t^2$ (2)

Solving (1) for t and substituting this expression in (2) gives:

$y = x \tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta} =x \tan \theta - \frac{gx^2}{2v^2}(1+ \tan^2 \theta)$ (Trigonometric identity)
$0 = \frac{-gx^2}{2v^2} \tan^2 \theta + x \tan \theta - \frac{gx^2}{2v^2} - y$

Let $p = \tan \theta$

$0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y$
$p = {\frac{-x\pm\sqrt{x^2-4(\frac{-gx^2}{2v^2})(\frac{-gx^2}{2v^2}-y)}}{2(\frac{-gx^2}{2v^2}) }}$ (Quadratic formula)
$\tan \theta = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}$
$\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)}$

Instead of a coordinate (x,y) it is required to hit a target at distance r and angle of elevation $\phi$ (polar coordinates), use the relationships $x = r \cos \phi$ and $y = r \sin \phi$ and substitute to get:

$\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi}\right)}$

## Catching balls

If a projectile, such as a baseball or cricket ball, travels in a parabolic path, with negligible air resistance, and if a player is positioned so as to catch it as it descends, he sees its angle of elevation increasing continuously throughout its flight. The tangent of the angle of elevation is proportional to the time since the ball was sent into the air, usually by being struck with a bat. Even when the ball is really descending, near the end of its flight, its angle of elevation seen by the player continues to increase. The player therefore sees it in line with a point ascending vertically from the batsman at constant speed. Finding the place from which the ball appears to rise steadily helps the player to position himself correctly to make the catch. If he is too close to the batsman who has hit the ball, it will appear to rise at an accelerating rate. If he is too far from the batsman, it will appear to slow rapidly, and then to descend.

Proof

Suppose the ball starts with a vertical component of velocity of $v_y$ upward, and a horizontal component of velocity of $v_x$ toward the player who wants to catch it. Its altitude above the ground is given by:

$h=v_y t-\frac{1}{2}gt^2,$ where $t$ is the time since the ball was hit.

The total time for the flight, until the ball is back down to the ground, $h=0$, is given by:

$\therefore T=\frac{2v_y}{g}.$

The horizontal component of the distance the ball travels from its starting point to time $t$ $(0 \le t \le T)$ is

$d=v_x t$

The total horizontal distance the ball travels from its starting point to the point where it is caught is:

$D=d(T)=\frac{2v _x v_y}{g}$

The horizontal component of the ball's distance from the catcher at time $t$ is:

$c=D-d= \frac{2v _x v_y}{g} - v_x t$

The tangent of the angle of elevation of the ball, as seen by the catcher, is:

$\tan(e)=\frac{h}{c}$
$=\frac{v_y t-\frac{1}{2}gt^2,}{\frac{2v_x v_y}{g}-v_x t}$
$=\frac{gt}{2 v_x}$

While the ball is in flight:

$\tan(e)=\left(\frac{g}{2 v_x}\right)t$

The bracket in this last expression is constant for a given flight. Therefore the tangent of the angle of elevation of the ball, as seen by the player who is properly positioned to catch it, is directly proportional to the time since the ball was hit.

## Trajectory of a projectile with air resistance

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

Air resistance will be taken to be in direct proportion to the velocity of the particle (i.e. $F_a \propto \vec{v}$). This is valid at low speed (low Reynolds number), and this is done so that the equations describing the particle's motion are easily solved. At higher speed (high Reynolds number) the force of air resistance is proportional to the square of the particle's velocity (see drag equation). Here, $v_0$,$v_x$ and $v_y$ 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. For the derivation only the case where $0^o \le \theta \le 180^o$ is considered. Again, the projectile is fired from the origin (0,0).

For this assumption, that air resistance may be taken to be in direct proportion to the velocity of the particle is not correct for a typical projectile in air with a velocity above a few tens of meters/second, and so this equation should not be applied to that situation.

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. $F_{air} = -kv$ (actually $F_{air} = -k v^2$ is more realistic, but not used here, to ensure an analytic solution,) is written due to the initial assumption of direct proportionality implies that the air resistance and the velocity differ only by a constant arbitrary factor with units of N*s/m.

As an example, say that when the velocity of the projectile is 4 m/s, the air resistance is 7 newtons (N). When the velocity is doubled to 8 m/s, the air resistance doubles to 14 N accordingly. In this case, k = 7/4 N x s/m. Note that k is needed in order to relate the air resistance and the velocity by an equal sign: otherwise, it would be stating incorrectly that the two are always equal in value (i.e. 1 m/s of velocity gives 1 N of force, 2 m/s gives 2 N etc.) which isn't always the case, and also it keeps the equation dimensionally correct (a force and a velocity cannot be equal to each other, e.g. m/s = N). As another quick example, Hooke's Law ($F = -kx$) describes the force produced by a spring when stretched a distance x from its resting position, and is another example of a direct proportion: k in this case has units N/m (in metric).

To show why k = 7/4 N·s/m above, first equate 4 m/s and 7 N:

$4 \ \mathrm{m}/\mathrm{s} = 7 \ \mathrm{N}$ (Incorrect)

$4 \ \mathrm{m}/\mathrm{s} \times (\frac{7}{4} \ \mathrm{N} \times \frac {\mathrm{s}}{\mathrm{m}})= 7 \ \mathrm{N}$ (Introduction of k)

$4 \ \mathrm{N} \times \frac{7}{4}= 7 \ \mathrm{N}$ ($\frac{\mathrm{s}}{\mathrm{m}} \times \frac{\mathrm{m}}{\mathrm{s}}$ cancels)

$7 \ \mathrm{N} = 7 \ \mathrm{N} (4 \times \frac{7}{4}) = 7$

For more on proportionality, see: Proportionality (mathematics)

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 $\Sigma F = -kv_x = ma_x$ and in the y direction $\Sigma F = -kv_y - mg = ma_y$.

This implies that: $a_x = \frac{-kv_x}{m} = \frac{dv_x}{dt}$ (1),

and

$a_y = \frac{1}{m}(-kv_y - mg) = \frac{-kv_y}{m} - g = \frac{dv_y}{dt}$ (2)
Solving (1) is an elementary differential equation, thus the steps leading to a unique solution for $v_x$ and, subsequently, $x$ will not be enumerated. Given the initial conditions $v_x = v_{xo}$ (where $v_{xo}$ is understood to be the x component of the initial velocity) and $s_x = 0$ for $t = 0$:

$v_x = v_{xo} e^{-\frac{k}{m}t}$ (1a)

$s_x = \frac{m}{k}v_{xo}(1-e^{-\frac{k}{m}t})$ (1b)

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 $v_y = v_{yo}$ and $s_y = 0$ when $t = 0$.

$\frac{dv_y}{dt} = \frac{-k}{m}v_y - g$ (2)

$\frac{dv_y}{dt} + \frac{k}{m}v_y = - g$ (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: $e^{\int \frac{k}{m} \, dt}$.

$e^{\frac{k}{m}t}(\frac{dv_y}{dt} + \frac{k}{m}v_y) = e^{\frac{k}{m}t}(-g)$ (2c)

$(e^{\frac{k}{m}t}v_y)^\prime = e^{\frac{k}{m}t}(-g)$ (2d)

$\int{(e^{\frac{k}{m}t}v_y)^\prime \,dt} = e^{\frac{k}{m}t}v_y = \int{ e^{\frac{k}{m}t}(-g) \, dt}$ (2e)

$e^{\frac{k}{m}t}v_y = \frac{m}{k} e^{\frac{k}{m}t}(-g) + C$(2f)

$v_y = \frac{-mg}{k} + Ce^{\frac{-k}{m}t}$ (2g)

And by integration we find:

$s_y = -\frac{mg}{k}t - \frac{m}{k}(v_{yo} + \frac{mg}{k})e^{-\frac{k}{m}t} + C$ (3)

Solving for our initial conditions:

$v_y(t) = -\frac{mg}{k} + (v_{yo} + \frac{mg}{k})e^{-\frac{k}{m}t}$ (2h)

$s_y(t) = -\frac{mg}{k}t - \frac{m}{k}(v_{yo} + \frac{mg}{k})e^{-\frac{k}{m}t} + \frac{m}{k}(v_{yo} + \frac{mg}{k})$ (3a)

With a bit of algebra to simplify (3a):
$s_y(t) = -\frac{mg}{k}t + \frac{m}{k}(v_{yo} + \frac{mg}{k})(1 - e^{-\frac{k}{m}t})$ (3b)

An example is given using values for the mass and terminal velocity for a baseball taken from [1].

m = 0.145 kg (5.1 oz)
v0 = 44.7 m/s (100 mph)
g = -9.81 m/s² (-32.2 ft/s²)
vt = -33.0 m/s (-73.8 mph)
$k =\frac{mg}{v_t} = \frac{(0.145 \mbox{ kg})(-9.81 \ \mathrm{m}/\mathrm{s}^2)}{-33.0 \ \mathrm{m}/\mathrm{s}} = 0.0431 \mbox{ kg}/\mbox{s} , \ \theta = 45^o$.

The green path (the lower path) is the path taken by the projectile modeled by the equations derived above, and the red path is taken by an idealized projectile, one that ignores air resistance altogether. (3.28 ft/m) Ignoring air resistance is not ideal in this scenario, as with no air resistance, a home run could be hit with 270 ft to spare. (The mechanics of pitching at 45 degrees notwithstanding.) And in most cases it's more accurate to assume $F_a \propto v^2$, meaning when velocity increases by a factor of $p$ the resistance increases by $p^2$. In the first example of proportionality, where the velocity was doubled to 8 m/s, the air resistance would instead be quadrupled ($2^2=4$) to 28 N: this only adds to the large amount of error in neglecting air resistance. For an analytic solution: Shouryya Ray solves one of Newton's puzzles

The more realistic trajectory, $F_{air} = -k v^2$ can also be calculated using modern method of calculus but this only add more complexity. Here (only in this condition), 'k' can also be calculated without any external experiment. i.e. k is also can be called Drag Factor

$k = \frac{1}{2}\rho\mathbf{A}C_D$

where

$\boldsymbol{\rho}$ is density of air

$\mathbf{A}$ is Frontal area of a projectile

$\boldsymbol{C_D}$ is called coefficient of drag, totally depends upon shape of a projectile

For a very very long projectile Skin Friction also plays important role due to very large amount of Total Surface Area which is also depends upon square function of speed of projectile.

As mentioned above, For the horizontal velocity after time $\mathbf{t}$ (projectile projecting at $\theta$ angle with the horizon)

$a_x = \frac{-kv^2_x}{m} = \frac{dv_x}{dt}$ (i)

$\frac{-kdt}{m} = \frac{dv_x}{v^2_x}$

$-\int \frac{k}{m} dt = \int v^{-2}_x dv_x$

$-\frac{kt}{m} = -\frac{1}{v_x} + c$

where $c$ is a constant of integration.

When $t=0$ , $v_x=v_{x_o}$

$c=\frac{1}{v_{x_o}}$

$\frac{kt}{m} = \frac{1}{v_{x}} - \frac{1}{v_{x_o}}$

$v_x = \frac{1}{\frac{1}{v_{x_0}}+\frac{kt}{m}}$

$v_x = \frac{1}{\frac{1}{v_0 \cos \theta}+\frac{kt}{m}}$ (ii)

Also for horizontal distance covered

$\int v_x dt = \int \left( \frac{1}{\frac{1}{v_0 \cos \theta}+\frac{kt}{m}} \right) dt$

$s_x(t) = \frac{m}{k}\ln (kt{v_0 \cos \theta}+m) + c$

When $t=0$ , $s_x=0$

$c = -\frac{m}{k}\ln (m)$

$s_x(t) = \frac{m}{k}(\ln (\frac{kt{v_0 \cos \theta}+m}{m}))$ (iii)

For the vertical component, it is much more tricky because while projectile is heading upward both gravity and air drag acts downward but when projectile is coming downward, gravity acts downward but air drag acts upward. So it has different solution for upward and downward direction.

While moving upward

$a_y= \frac{-kv^2_y}{m} - g = \frac{dv_y}{dt}$

This can be solved to get

$v_y(t) = \sqrt{\frac{mg}{k}} \tan (c - \sqrt{\frac{gk}{m}} t)$

When $t=0$ , $v_y(t)=v_{y_o}=v_0 \sin \theta$

$c = \arctan(\sqrt{\frac{k}{mg}} v_{y_o})$

$v_y(t) = \sqrt{\frac{mg}{k}} \tan (\arctan(\sqrt{\frac{k}{mg}} v_{y_o}) - \sqrt{\frac{gk}{m}} t)$ (iv)

Also for vertical height covered

$\int ds_y = \int (\sqrt{\frac{mg}{k}} \tan (\arctan(\sqrt{\frac{k}{mg}} v_{y_o}) - \sqrt{\frac{gk}{m}} t)) dt$

$s_y(t) = \frac{m}{k} \ln( \cos( \sqrt{ \frac {gk}{m} } t - \arctan( \sqrt{ \frac {k}{mg}} v_{y_o} ) ) ) + c$

When $t=0$ , $s_y(t)=0$

$c = - \frac{m}{k} \ln( \cos( \arctan( \sqrt{ \frac {k}{mg}} v_{y_o} ) ) )$

$s_y(t) = \frac{m}{k} \ln( \cos( \sqrt{ \frac {gk}{m} } t - \arctan( \sqrt{ \frac {k}{mg}} v_{y_o}))) - \frac{m}{k} \ln( \cos( \arctan( \sqrt{ \frac {k}{mg}} v_{y_o} ) ) )$ (v)

So if $\mathbf{h}$ is the maximum height and $\mathbf{t_{ascend}}$ is the required time to reach there then

$t_{ascend} = \sqrt{\frac{m}{gk}} \arctan( \sqrt{ \frac {k}{mg} } v_{y_o})$ (vi)

$h = - \frac{m}{k} \ln( \cos( \arctan( \sqrt{ \frac {k}{mg}} v_{y_o} ) ) )$ (vii)

Similarly for when the projectile is coming downward. Here $\mathbf{t}$ is the descending time only. ie. $\mathbf{t}$ is counted only after reaching the maximum height.

$a_y= \frac{kv^2_y}{m} - g = \frac{dv_y}{dt}$

This too can be solved to get

$v_y(t) = - \sqrt{\frac{mg}{k}} \tanh (c + \sqrt{\frac{gk}{m}} t)$

When $t=0$ , $v_y(t)=0$

$c = 0$

$v_y(t) = - \sqrt{\frac{mg}{k}} \tanh (\sqrt{\frac{gk}{m}} t)$ (viii)

And if $\mathbf{s_y(t)}$ is the height above the surface at a particular time after $\mathbf{t_{ascend}}$

$s_y(t) = - \frac{m}{k} \ln( \cosh( \sqrt{\frac{gk}{m}} t)) + c$

When $t=0$ , $s_y(t)=h$, where $\mathbf{h}$ is the maximum height

$c = h$ (ix)

$s_y(t) = h - \frac{m}{k} \ln( \cosh( \sqrt{\frac{gk}{m}} t))$ (ix)

So if $\mathbf{t_{descend}}$ is the required time to hit the ground

$t_{descend} = \sqrt{\frac{m}{gk}} \operatorname{arccosh}\,(e^{\frac{hk}{m}})$ (x)

$t_{descend} = \sqrt{\frac{m}{gk}} \operatorname{arccosh}\,(\sqrt{1+\frac{kv^2_{y_o}}{gm}})$ (x-a)

Finally, the velocity at any instant $\mathbf{v(t)}$, Total Time of flight $\mathbf{T}$, Range $\mathbf{R}$ and speed at which it hit the ground $\mathbf{V_{hit}}$ is

$v(t) = \sqrt{v^2_x(t) + v^2_y(t)}$ (xi)

$T= t_{ascend} + t_{descend}$ (xii)

$R = \frac{m}{k}(\ln (\frac{kT{v_0 \cos \theta}+m}{m}))$ (xiii)

$V_{hit} = \sqrt{(\frac{1}{\frac{1}{v_0 \cos \theta}+\frac{kT}{m}})^2 + \frac{mg}{k}( \tanh( \operatorname{arccosh}\,(e^\frac{hk}{m})))^2}$ (xiv)

If the projectile is projected for a very long range then maximum height also increase then the density of air must be changed from instant to instant, meaning changing 'k', this only add further complication. Within the Troposphere, change in density of air can be predicted with only negligible amount of error.

$\rho = \frac{MP_o}{R(T_o-Lh)} \times [1 - \frac{Lh}{T_o}]^\frac{gM}{RL}$

Where

$\mathbf{L}$ is Temperature Lapse rate

$\mathbf{P_o}$ is Pressure at sea level

$\mathbf{g}$ is mean acceleration due to gravity

$\mathbf{T_o}$ is Absolute Temperature at sea level i.e. in K

$\mathbf{h}$ is height above sea level

$\mathbf{M}$ is Molar mass of Air

$\mathbf{R}$ is Gas constant