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Trajectory Of A Projectile
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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


Distance traveled

The total horizontal distance d traveled.

: d = rac{v \cos heta}{g} \left( v \sin heta + \sqrt{v^2 \sin^2 heta + 2gy_0} ight)

As a special case, the distance is given by

: d = rac{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 = rac{d}{v \cos heta} = rac{v \sin heta + \sqrt{(v \sin heta)^2 + 2gy_0}}{g}

As above, this expression can be reduced to

: t = rac{\sqrt{2} \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.

: \sin(2 heta) = rac{gd}{v^2}

: heta = rac{1}{2} \sin^{-1} \left( rac{gd}{v^2} ight)


CONDITIONS AT AN ARBITRARY DISTANCE <VAR>X</VAR>


Height at x

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

: y = x an heta - rac {gx^2}{2(v\cos heta)^2} + y_0 .


Velocity at x

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

: s = \sqrt{v^2 + 2gx an heta + rac{g^2}{\cos^2 heta}} .


Derivation

The magnitude s of the velocity is given by

: s = \sqrt{V_x^2 + V_y^2} ,

where Vx and Vy 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 vi = v sin θ, a = g, and t = rac{x}{v \cos heta}. (The latter is found by taking x = (v cos θ) t and solving for t.) Then,

: V_y = v \sin heta + rac{gx}{v \cos heta}

and

: s = \sqrt{(v \cos heta)^2 + \left(v \sin heta + rac{gx}{v \cos heta} ight)^2} .

The formula above is found by simplifying.


ANGLE <MATH> HETA</MATH> REQUIRED TO HIT COORDINATE (<VAR>X</VAR>,<VAR>Y</VAR>)

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 heta are:

: heta = an^{-1}{\left( rac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} ight)}

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 elemtary formulae relating to projectile motion:

x = v t \cos heta , t = rac{x}{v \cos heta} (1)

y = vt \sin heta - rac{1}{2} g t^2 (2)

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

y = x an heta - rac{gx^2}{2v^2 \cos^2 heta} (2a)

y = x an heta - rac{gx^2 \sec^2 heta}{2v^2} (2b) (Trigonometric identity)

y =x an heta - rac{gx^2}{2v^2}(1+ an^2 heta) (2c) (Trigonometric identity)

0 = rac{-gx^2}{2v^2} an^2 heta + x an heta - rac{gx^2}{2v^2} - y (2d) (Algebra)

Let p = an heta

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

p = { rac{-x\pm\sqrt{x^2-4( rac{-gx^2}{2v^2})( rac{-gx^2}{2v^2}-y)}}{2( rac{-gx^2}{2v^2}) }} (2f) ( Quadratic Formula )

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

an heta = rac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2g) (Substitution)

heta = an^{-1}{\left( rac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} ight)} (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:

heta = an^{-1}{\left( rac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi} ight)}


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 ec{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 heta \le 180^o is considered. Again, the projectile is fired from the origin (0,0).

Above is a (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 \ \mathrm{m}/\mathrm{s} = 7 \ \mathrm{N} (Incorrect)

4 \ \mathrm{m}/\mathrm{s} imes ( rac{7}{4} \ \mathrm{N} imes rac {\mathrm{s}}{\mathrm{m}})= 7 \ \mathrm{N} (Introduction of k)

4 \ \mathrm{N} imes rac{7}{4}= 7 \ \mathrm{N} ( rac{\mathrm{s}}{\mathrm{m}} imes rac{\mathrm{m}}{\mathrm{s}} cancels)

7 \ \mathrm{N} = 7 \ \mathrm{N} (4 imes rac{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 (\Sigma F = ma) for both the x and y components:

m rac{dv_x}{dt}(=a_x)= -kv_x (1)

m rac{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 = rac{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 \cos heta and x=0 for t=0, these solutions are:

v_x = v_0 e^{- rac{kt}{m}}\cos heta (1a)

x = rac{mv_0\cos heta}{k}(1-e^{- rac{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 \sin heta and y=0 for t=0

m rac{dv_y}{dt} = -kv_y+mg (2)

rac{dv_y}{-kv_y+mg}= rac{1}{m}dt (2a) (Algebra)

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

- rac{1}{k}ln ert-kv_y+mg ert= rac{t}{m}+C (2c) (Result of integration)

- rac{1}{k}ln ert-kv_y+mg ert= rac{t}{m}- rac{1}{k}ln ert-kv_0 \sin heta +mg ert (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 \le heta \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 \sin heta \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( rac{-kv_y+mg}{-kv_0 \sin heta+mg} ight)=- rac{kt}{m} (2e) (Algebra)

rac{-kv_y+mg}{-kv_0 \sin heta+mg}=e^{- rac{kt}{m}} (2f) (Exponentiation, or anti-logarithm)

v_y = - rac{mg}{k}e^{- rac{kt}{m}}+v_0e^{- rac{kt}{m}}\sin heta+ rac{mg}{k} (2g) (Algebra)

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

\int{dy}=\int{(- rac{mg}{k}e^{- rac{kt}{m}}+v_0e^{- rac{kt}{m}}\sin heta+ rac{mg}{k})dt} (2i) (Integration)

y = rac{m^2g}{k^2}e^{- rac{kt}{m}}- rac{mv_0 \sin heta}{k}e^{- rac{kt}{m}}+ rac{mgt}{k}+C (Result of integration) (2j)

y = rac{mv_0\sin heta}{k}(1-e^{- rac{kt}{m}})- rac{m^2g}{k^2}(e^{- rac{kt}{m}}+ rac{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 ightarrow\infty, we see that e^{- rac{kt}{m}} ightarrow 0, and all disapprears but the rac{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 = rac{mg}{v_t}

Also worth noting is that if we take the limit as t ightarrow\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}= rac{mv_0\cos heta}{k}

Also, for interest the solutions for v_y and y in the case where -180^o \ge heta \ge 0^o are:

v_y = rac{mg}{k}e^{- rac{kt}{m}}-v_0e^{- rac{kt}{m}}\sin heta+ rac{mg}{k}

y = rac{mv_0\sin heta}{k}(1-e^{- rac{kt}{m}})+ rac{m^2g}{k^2}(1+ rac{kt}{m}-e^{- rac{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 {Link without Title} .
m

v

g

v

:k = rac{mg}{v_t} = rac{(0.145 \mbox{ kg})(-9.81 \ \mathrm{m}/\mathrm{s}^2)}{-33.0 \ \mathrm{m}/\mathrm{s}} = 0.0431 \mbox{ kg}/\mbox{s} , \ heta = 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 \propto ec{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.