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Trajectory & Range of a Projectile

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.

In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range.

This article provides a list of methods for calculating the trajectory and range of a projectile under the influence of Earth's gravity.

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

1 Conditions at the final position of the projectile
2 Conditions at an arbitrary distance x
- 2.1 Height at x
- 2.2 Velocity at x
  - 2.2.1 Derivation
3 Angle θ required to hit coordinate (x,y)
4 Other meaning
4 Range of a Projectile
5 External links
6 Trajectory of a projectile with air resistance

Conditions at the final position of the projectile

Distance traveled

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 distance traveled is:

$d = \frac{v_0^2 \sin(2 \theta)}{g}$

As a special case, the distance is given by

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

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

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 (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v.

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

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

External links

Spreadsheet to calculate distance and time of flight with your own values to understand the equation

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.

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 + a t

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

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

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

Each root of the equation corresponds to the two possible launch angles so long as both roots aren't imaginary, in which case the initial 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:

$x = v t \cos \theta , t = \frac{x}{v \cos \theta}$ (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 $p = tanθ$

$0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y$ (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 $φ$ (polar coordinates), use the relationships $x = r cosφ$ and $y = r sinφ$ and substitute to get:

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

Other meaning

The US-DOD and NATO definition of a ballistic trajectory is a trajectory traced after the propulsive force is terminated and the body is acted upon only by gravity and aerodynamic drag.

External links

Java applet of projectile motion

Trajectory of a projectile with air resistance

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. $F_a \propto \vec{v}$ . 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 $0^o \le \theta \le 180^o$ is considered. Again, the projectile is fired from the origin (0,0).

Note: This assumption the force 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.

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 $F a i r = - k v$ 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 ( $F = - k x$ ) 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)

To derive relationships to represent the motion of the particle, we first apply Newton's Second Law ( $Σ F = m a$ ) for both the x and y components:

$m \frac{dv_x}{dt}(=a_x)= -kv_x$ (1)

$m \frac{dv_y}{dt}(=a_y) = -kv_y+mg$ (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 ( $a = \frac{dv}{dt}$ ). 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 $v x$ = $v 0 cosθ$ and $x = 0$ for $t = 0$ , these solutions are:

$v_x = v_0 e^{-\frac{kt}{m}}\cos \theta$ (1a)

$x = \frac{mv_0\cos\theta}{k}(1-e^{-\frac{kt}{m}})$ (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 $v y = v 0 sinθ$ and $y = 0$ for t=0

$m \frac{dv_y}{dt} = -kv_y+mg$ (2)

$\frac{dv_y}{-kv_y+mg}=\frac{1}{m}dt$ (2a)

$\int{\frac{dv_y}{-kv_y+mg}}=\int{\frac{1}{m}dt}$ (2b)

$-\frac{1}{k}ln\vert-kv_y+mg\vert=\frac{t}{m}+C$ (2c)

$-\frac{1}{k}ln\vert-kv_y+mg\vert=\frac{t}{m}-\frac{1}{k}ln\vert-kv_0 \sin \theta +mg\vert$ (2d) (Substitution 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 (multiply the number of possible cases by 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 $- k v y$ can never exceed $m g$ (otherwise air resistance would cause the object to accelerate upward against gravity, this can only happen if the projectile is initially fired with a negative y-component velocity greater than the object's terminal velocity). And because we are only considering the case where $0^o \le \theta \le 180^o$ , the right-hand member within the absolute value signs is always negative, since $v 0$ can never exceed $v y$ (and thus $- k v 0$ cannot exceed mg) and $\sin \theta \ge 0$ . 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.

$ln\left(\frac{-kv_y+mg}{-kv_0 \sin \theta+mg}\right)=-\frac{kt}{m}$ (2e)

$\frac{-kv_y+mg}{-kv_0 \sin \theta+mg}=e^{-\frac{kt}{m}}$ (2f)

$v_y = -\frac{mg}{k}e^{-\frac{kt}{m}}+v_0e^{-\frac{kt}{m}}\sin \theta+\frac{mg}{k}$ (2g)

$\frac{dy}{dt} = -\frac{mg}{k}e^{-\frac{kt}{m}}+v_0e^{-\frac{kt}{m}}\sin \theta+\frac{mg}{k}$ (2h) (Substitute $v_y = \frac{dy}{dt}$ )

$\int{dy}=\int{(-\frac{mg}{k}e^{-\frac{kt}{m}}+v_0e^{-\frac{kt}{m}}\sin \theta+\frac{mg}{k})dt}$ (2i)

$y = \frac{m^2g}{k^2}e^{-\frac{kt}{m}}-\frac{mv_0 \sin \theta}{k}e^{-\frac{kt}{m}}+\frac{mgt}{k}+C$ (2j)

$y =\frac{mv_0\sin\theta}{k}(1-e^{-\frac{kt}{m}})+\frac{m^2g}{k^2}(e^{-\frac{kt}{m}}+\frac{kt}{m}-1)$ (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 $t\rightarrow\infty$ , we see that $e^{-\frac{kt}{m}}\rightarrow 0$ , and all disappears but the $\frac{mg}{k}$ 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:

$k = \frac{mg}{v_t}$

Also worth noting is that if we take the limit as $t\rightarrow\infty$ 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:

$x_{max}=\frac{mv_0\cos\theta}{k}$

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

$v_y =\frac{mg}{k}e^{-\frac{kt}{m}}-v_0e^{-\frac{kt}{m}}\sin \theta+\frac{mg}{k}$

$y =\frac{mv_0\sin\theta}{k}(1-e^{-\frac{kt}{m}})+\frac{m^2g}{k^2}(1+\frac{kt}{m}-e^{-\frac{kt}{m}})$ (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 [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)

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

(This graph was produced using "GraphCalc")

The red path is the path taken by our projectile modeled 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 $F_a \propto \vec{v}^2$ , 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 ( $22 = 4$ ) to 28 N: this only adds to the large amount of error in neglecting air resistance.

Range of a Projectile

The path of this projectile launched from a height y0 has a range d.

The path of this projectile launched from a height y₀ has a range d.

In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range. As in Trajectory of a projectile, we will use:

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 travelled by the projectile

When neglecting air resistance, the range of a projectile will be

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

If (y₀) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify to

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

Derivations

Flat Ground

First we examine the case where (y₀) is zero. The horizontal position (x(t)) of the projectile is

$x(t) = \frac{}{} v\cos \left(\theta\right) t$

In the vertical direction

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

We are interested in the time when the projectile returns to the same height it originated at, thus

$0 = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

By applying the quadratic formula

$\frac{} {}t = 0$

$t = \frac{2 v \sin \theta} {g}$

The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation yields

$x = \frac {2 v^2 \cos \left(\theta\right) \sin \left(\theta\right)} {g}$

Applying the trigonometric identity

$\sin(2x) = 2 \sin (x) \cos(x) \$

allows us to simplify the solution to

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

Note that when (θ) is 45°, the solution becomes

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

Uneven Ground

Now we will allow (y₀) to be nonzero. Our equations of motion are now

$x(t) = \frac{}{} v\cos \left(\theta\right) t$

and

$y(t) = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with)

$0 = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation

$t = \frac {v \sin \theta} {g} \pm \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}$

The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is

$t = \frac {v \sin \theta} {g} + \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}$