In physics, the ballistic trajectory of a projectile is the path that a thrown object will take under the action of gravity, neglecting all other forces, such as friction from air resistance, or propulsion. This article provides a list of methods for calculating the trajectory of a projectile under the influence of Earth's gravity.

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

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

Conditions at the final position of the projectile

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Distance traveled

The total horizontal distance d traveled.

$d\; =\; \backslash frac\{v\; \backslash cos\; \backslash theta\}\{g\}\; \backslash left(\; v\; \backslash sin\; \backslash theta\; +\; \backslash sqrt\{v^2\; \backslash sin^2\; \backslash theta\; +\; 2gy\_0\}\; \backslash right)$

As a special case, the distance is given by

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

when the angle θ is 45° and the initial height y₀ is 0.

Time of flight

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

$t\; =\; \backslash frac\{d\}\{v\; \backslash cos\backslash theta\}\; =\; \backslash frac\{v\; \backslash sin\; \backslash theta\; +\; \backslash sqrt\{(v\; \backslash sin\; \backslash theta)^2\; +\; 2gy\_0\}\}\{g\}$

As above, this expression can be reduced to

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

if θ is 45° and y₀ is 0.

Angle of reach

The "angle of reach" (not quite a scientific term) is the angle θ at which a projectile must be launched in order to go a distance d, given the initial velocity v.

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

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

Conditions at an arbitrary distance x

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Height at x

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

$y\; =\; x\; \backslash tan\; \backslash theta\; -\; \backslash frac\; \{gx^2\}\{2(v\backslash cos\backslash theta)^2\}\; +\; y\_0$.

Velocity at x

The magnitude s of the velocity of the projectile at distance x is given by

$s\; =\; \backslash sqrt\{v^2\; +\; 2gx\; \backslash tan\; \backslash theta\; +\; \backslash frac\{g^2\}\{\backslash cos^2\; \backslash theta\}\}$.

Derivation

The magnitude s of the velocity is given by

$s\; =\; \backslash sqrt\{V\_x^2\; +\; V\_y^2\}$,

where V_x and V_y are the instantaneous velocities in the x- and y-directions, respectively.

We can see that 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 v_i = v sin θ, a = g, and $t\; =\; \backslash frac\{x\}\{v\; \backslash cos\; \backslash theta\}$. (The latter is found by taking x = (v cos θ) t and solving for t.) Then,

$V\_y\; =\; v\; \backslash sin\; \backslash theta\; +\; \backslash frac\{gx\}\{v\; \backslash cos\; \backslash theta\}$

and

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

The formula above is found by simplifying.

Angle $\backslash theta$ required to hit coordinate (x,y)

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To hit a target at range x and altitude y when fired from (0,0) and with intitial velocity v the required angle(s) of launch $\backslash theta$ are:

$\backslash theta\; =\; \backslash tan^\{-1\}\{\backslash left(\backslash frac\{v^2\backslash pm\backslash sqrt\{v^4-g(gx^2+2yv^2)\}\}\{gx\}\backslash right)\}$

Each root of the equation cooresponds to the two possible launch angles so long as both roots aren't imaginary, in which case the intitial velocity is not great enough to reach the point (x,y) you have selected. The greatest feature of this formula is that it allows you to find the angle of launch needed without the restriction of y = 0.

Derivation

First, we call upon two elementary formulae relating to projectile motion:

$x\; =\; v\; t\; \backslash cos\; \backslash theta\; ,\; t\; =\; \backslash frac\{x\}\{v\; \backslash cos\; \backslash theta\}$ (1)

$y\; =\; vt\; \backslash sin\; \backslash theta\; -\; \backslash frac\{1\}\{2\}\; g\; t^2$ (2)

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

$y\; =\; x\; \backslash tan\; \backslash theta\; -\; \backslash frac\{gx^2\}\{2v^2\; \backslash cos^2\; \backslash theta\}$ (2a)

$y\; =\; x\; \backslash tan\; \backslash theta\; -\; \backslash frac\{gx^2\; \backslash sec^2\; \backslash theta\}\{2v^2\}$ (2b) (Trigonometric identity)

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

$0\; =\; \backslash frac\{-gx^2\}\{2v^2\}\; \backslash tan^2\; \backslash theta\; +\; x\; \backslash tan\; \backslash theta\; -\; \backslash frac\{gx^2\}\{2v^2\}\; -\; y$ (2d) (Algebra)

Let $p\; =\; \backslash tan\; \backslash theta$

$0\; =\; \backslash frac\{-gx^2\}\{2v^2\}\; p^2\; +\; xp\; -\; \backslash frac\{gx^2\}\{2v^2\}\; -\; y$ (2e) (Substitution)

$p\; =\; \{\backslash frac\{-x\backslash pm\backslash sqrt\{x^2-4(\backslash frac\{-gx^2\}\{2v^2\})(\backslash frac\{-gx^2\}\{2v^2\}-y)\}\}\{2(\backslash frac\{-gx^2\}\{2v^2\})\; \}\}$ (2f) (Quadratic formula)

$p\; =\; \backslash frac\{v^2\backslash pm\backslash sqrt\{v^4-g(gx^2+2yv^2)\}\}\{gx\}$ (2f) (Algebra)

$\backslash tan\; \backslash theta\; =\; \backslash frac\{v^2\backslash pm\backslash sqrt\{v^4-g(gx^2+2yv^2)\}\}\{gx\}$ (2g) (Substitution)

$\backslash theta\; =\; \backslash tan^\{-1\}\{\backslash left(\backslash frac\{v^2\backslash pm\backslash sqrt\{v^4-g(gx^2+2yv^2)\}\}\{gx\}\backslash right)\}$ (2h) (Algebra)

Also, if instead of a coordinate (x,y) you're interested in hitting a target at distance r and angle of elevation $\backslash phi$ (polar coordinates), use the relationships $x\; =\; r\; \backslash cos\; \backslash phi$ and $y\; =\; r\; \backslash sin\; \backslash phi$ and substitute to get:

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

Trajectory of a projectile with air resistance

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Note: This section considers the case where the force of air resistance may be taken to be in direct proportion to the velocity of the particle i.e. $F\_a\; \backslash propto\; \backslash vec\{v\}$. Also, $v\_0$,$v\_x$ 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 $0^o\; \backslash le\; \backslash theta\; \backslash le\; 180^o$ is considered. Again, the projectile is fired from the origin (0,0).

Above is a free body diagram (not to scale) for a projectile that experiences air resistance and the effects of gravity (the dashed vectors are the x and y components of velocity and air resistance). Here we assume that the air resistance is in the direction opposite of the projectile's velocity. We can write $F\_\{air\}\; =\; -kv$ because our initial assumption of direct proportionality implies that the air resistance and the velocity differ only by a constant arbitrary factor, and that when v is increased by a factor of, say, p, the air resistance increases by a factor of p also. As an example, say that when the velocity of the projectile is 4 m/s, the air resistance is 7 newtons (N). When we double the velocity to 8 m/s, the air resistance doubles to 14 N accordingly. In this case, k = 7/4 N x s/m. Note that we need k in order to relate the air resistance and the velocity by an equal sign: otherwise, we'd 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 (we can't have a force and a velocity 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 streched 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:

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

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

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

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

For more on proportionality, see: Proportionality (mathematics)

To derive relationships to represent the motion of the particle, we first apply Newton's Second Law ($\backslash Sigma\; F\; =\; ma$) for both the x and y components:

$m\; \backslash frac\{dv\_x\}\{dt\}(=a\_x)=\; -kv\_x$ (1)

$m\; \backslash frac\{dv\_y\}\{dt\}(=a\_y)\; =\; -kv\_y+mg$ (2) (The mg term is positive because the value of g is already negative and subtracting it would result in a positive number.)

Note that acceleration is just the derivative of velocity with respect to time ($a\; =\; \backslash frac\{dv\}\{dt\}$). Solving (1) is an elementary problem in solving differential equations and the solution for $v\_x$ and, subsequently, x will not be given proof. For initial conditions $v\_x$= $v\_0\; \backslash cos\; \backslash theta$ and $x=0$ for $t=0$, these solutions are:

$v\_x\; =\; v\_0\; e^\{-\backslash frac\{kt\}\{m\}\}\backslash cos\; \backslash theta$ (1a)

$x\; =\; \backslash frac\{mv\_0\backslash cos\backslash theta\}\{k\}(1-e^\{-\backslash frac\{kt\}\{m\}\})$ (1b)

(2) will be solved here for interest. In fact, (1) is solved in much the same way. Note that in this case we use the initial conditions $v\_y\; =\; v\_0\; \backslash sin\backslash theta$ and $y=0$ for t=0

$m\; \backslash frac\{dv\_y\}\{dt\}\; =\; -kv\_y+mg$ (2)

$\backslash frac\{dv\_y\}\{-kv\_y+mg\}=\backslash frac\{1\}\{m\}dt$ (2a) (Algebra)

$\backslash int\{\backslash frac\{dv\_y\}\{-kv\_y+mg\}\}=\backslash int\{\backslash frac\{1\}\{m\}dt\}$ (2b) (Integration)

$-\backslash frac\{1\}\{k\}ln\backslash vert-kv\_y+mg\backslash vert=\backslash frac\{t\}\{m\}+C$ (2c) (Result of integration)

$-\backslash frac\{1\}\{k\}ln\backslash vert-kv\_y+mg\backslash vert=\backslash frac\{t\}\{m\}-\backslash frac\{1\}\{k\}ln\backslash vert-kv\_0\; \backslash sin\; \backslash theta\; +mg\backslash vert$ (2d) (Subtitution of initial values, solved for C, substitution with the result for C)

At this point because we have the absolute value function in our equation, we would normally have to solve four different cases (two for each term in the absolute value signs). However, the absolute value term in the left-hand member is always negative, because the term $-kv\_y$ can never exceed $mg$ (otherwise air resistance would cause the object to move upward against gravity). And because we are only considering the case where $0^o\; \backslash le\; \backslash theta\; \backslash le\; 180^o$, the right-hand member within the absolute value signs is always negative, since $v\_0$ can never exceed $v\_y$ (and thus $-kv\_0$ cannot exceed mg) and $\backslash sin\; \backslash theta\; \backslash ge\; 0$. Thus when we go to combine the two terms in the next step, the quotient that appears is always positive, and the absolute value signs can be omitted.

$ln\backslash left(\backslash frac\{-kv\_y+mg\}\{-kv\_0\; \backslash sin\; \backslash theta+mg\}\backslash right)=-\backslash frac\{kt\}\{m\}$ (2e) (Algebra)

$\backslash frac\{-kv\_y+mg\}\{-kv\_0\; \backslash sin\; \backslash theta+mg\}=e^\{-\backslash frac\{kt\}\{m\}\}$ (2f) (Exponentiation, or anti-logarithm)

$v\_y\; =\; -\backslash frac\{mg\}\{k\}e^\{-\backslash frac\{kt\}\{m\}\}+v\_0e^\{-\backslash frac\{kt\}\{m\}\}\backslash sin\; \backslash theta+\backslash frac\{mg\}\{k\}$ (2g) (Algebra)

$\backslash frac\{dy\}\{dt\}\; =\; -\backslash frac\{mg\}\{k\}e^\{-\backslash frac\{kt\}\{m\}\}+v\_0e^\{-\backslash frac\{kt\}\{m\}\}\backslash sin\; \backslash theta+\backslash frac\{mg\}\{k\}$ (2h) (Substitute $v\_y\; =\; \backslash frac\{dy\}\{dt\}$)

$\backslash int\{dy\}=\backslash int\{(-\backslash frac\{mg\}\{k\}e^\{-\backslash frac\{kt\}\{m\}\}+v\_0e^\{-\backslash frac\{kt\}\{m\}\}\backslash sin\; \backslash theta+\backslash frac\{mg\}\{k\})dt\}$ (2i) (Integration)

$y\; =\; \backslash frac\{m^2g\}\{k^2\}e^\{-\backslash frac\{kt\}\{m\}\}-\backslash frac\{mv\_0\; \backslash sin\; \backslash theta\}\{k\}e^\{-\backslash frac\{kt\}\{m\}\}+\backslash frac\{mgt\}\{k\}+C$ (Result of integration) (2j)

$y\; =\backslash frac\{mv\_0\backslash sin\backslash theta\}\{k\}(1-e^\{-\backslash frac\{kt\}\{m\}\})-\backslash frac\{m^2g\}\{k^2\}(e^\{-\backslash frac\{kt\}\{m\}\}+\backslash frac\{kt\}\{m\}-1)$ (2k) (Substitution of initial values, solved for C, substitution with the result for C, and factorization)

Looking back at the equation for the y-component for velocity (2g), we can find a good way of calculating a numerical value for k. If we take the limit of (2g) as $t\backslash rightarrow\backslash infty$, we see that $e^\{-\backslash frac\{kt\}\{m\}\}\backslash rightarrow\; 0$, and all disapprears but the $\backslash frac\{mg\}\{k\}$term. $v\_y$ is also affected by time, but it does not grow indefinitely: the projectile will approach its terminal velocity (in the y direction) as time passes indefinitely, which we'll call $v\_t$. Using this in (2g) gives us:

$k\; =\; \backslash frac\{mg\}\{v\_t\}$

Also worth noting is that if we take the limit as $t\backslash rightarrow\backslash infty$ in equation (1b) we see that there is a maximum value that can be reached by x (if the projectile doesn't hit the ground first). This is given by:

$x\_\{max\}=\backslash frac\{mv\_0\backslash cos\backslash theta\}\{k\}$

Also, for interest the solutions for $v\_y$ and $y$ in the case where $-180^o\; \backslash ge\; \backslash theta\; \backslash ge\; 0^o$ are:

$v\_y\; =\backslash frac\{mg\}\{k\}e^\{-\backslash frac\{kt\}\{m\}\}-v\_0e^\{-\backslash frac\{kt\}\{m\}\}\backslash sin\; \backslash theta+\backslash frac\{mg\}\{k\}$

$y\; =\backslash frac\{mv\_0\backslash sin\backslash theta\}\{k\}(1-e^\{-\backslash frac\{kt\}\{m\}\})+\backslash frac\{m^2g\}\{k^2\}(1+\backslash frac\{kt\}\{m\}-e^\{-\backslash frac\{kt\}\{m\}\})$ (The solutions for $v\_x$ and $x$ are not affected.)

An example is given using values for the mass and terminal velocity for a baseball taken from ^*.

m = 0.145 kg (5.1 oz)

v₀ = 44.7 m/s (100 mi/h)

g = -9.81 m/s² (-32.2 ft/s²)

v_t = -33.0 m/s (-73.8 mi/h)

$k\; =\backslash frac\{mg\}\{v\_t\}\; =\; \backslash frac\{(0.145\; \backslash mbox\{\; kg\})(-9.81\; \backslash \; \backslash mathrm\{m\}/\backslash mathrm\{s\}^2)\}\{-33.0\; \backslash \; \backslash mathrm\{m\}/\backslash mathrm\{s\}\}\; =\; 0.0431\; \backslash mbox\{\; kg\}/\backslash mbox\{s\}\; ,\; \backslash \; \backslash theta\; =\; 45^o$.

(This graph was produced using "GraphCalc")

The red path is the path taken by our projectile modelled by our equations derived above, and the green path is taken by an idealized projectile, one that ignores air resistance altogether. (For those of you who'd prefer those numbers in feet, the conversion factor is 3.28 ft/m) Turns out ignoring air resistance isn't a very good idea (in this case at least): without it a pitcher could throw a home run with 270 ft to spare! (The mechanics of pitching at 45 degrees notwithstanding.) And in some cases it's more accurate to assume $F\_a\; \backslash propto\; \backslash vec\{v\}^2$, meaning when air resistance increases by a factor of p the resistance increases by $p^2$. To go back to the first example of proportionality, when we doubled the velocity 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 negelcting air resistance.

Ballistics

Trayectoria balística