Trajectory of a projectile

In physics, the ballistic trajectory of a projectile is the path that a thrown or launched projectile will take under the action of gravity, neglecting all other forces, such as friction from air resistance, 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.

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 velocity 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'), "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.

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

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

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

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 V_x and V_y 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 v_i = 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) 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, two elementary formulae are called upon relating to projectile motion:

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

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

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

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

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

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

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

Let $p = \tan \theta$

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

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

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

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

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

Also, if instead of a coordinate (x,y) you're interested in hitting a target at distance r and angle of elevation $\phi$ (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 as if it were ascending vertically 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 $v,$ upward, and a horizontal component of velocity of $h$ toward the player who wants to catch it. Its altitude above the ground is given by:

$a=vt-\frac{1}{2}gt^2,$ where $t$ is the time since the ball was hit, and $g$ is the acceleration due to gravity.

The total time for the flight, until the ball is back down to the ground, from which it started, is given by:

$a=0$

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

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

$d=h(T-t) = \frac{2hv}{g}-ht$

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

$\tan(e)=\frac{a}{d}$

$=\frac{vt-\frac{gt^2}{2}}{\frac{2hv}{g}-ht}$

$=\frac{2gvt-g^2t^2}{4hv-2ght}$

$=\frac{gt(2v-gt)}{2h(2v-gt)}$

The quantity in the brackets will not be zero except when the ball is on the ground, therefore, while the ball is in flight:

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

The bracket in this last expression is constant for a given flight of the ball. 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)

v_t = -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 red path (the lower path) is the path taken by the projectile modeled by the equations derived above, and the green 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 $F_a \propto \vec{v}^2$, meaning when air resistance 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

See also

• Ballistic coefficient
• Range of a projectile
• Trajectory

In fluid dynamics, drag (sometimes called air resistance or fluid resistance) refers to forces which act on a solid object in the direction of the relative fluid flow velocity.[1] [2] [3] [4] Unlike other resistive forces such as dry friction, which is nearly independent of velocity, drag forces depend on velocity.[5] Drag forces always decrease fluid velocity relative to the solid object in the fluid's path.

External links

• Open Source Physics computer model
• Spreadsheet to calculate distance and time of flight
• Java applet of projectile motion
• [2]