# Displacement (vector)

Displacement versus distance traveled along a path.
A displacement is the shortest distance from the initial to the final position of a point P[1]. Thus, it is the length of an imaginary straight path, typically distinct from the path actually travelled by P. A displacement vector represents the length and direction of that imaginary straight path.
A position vector expresses the position of a point P in space in terms of a displacement from an arbitrary reference point O (typically the origin of a coordinate system). Namely, it indicates both the distance and direction of an imaginary motion along a straight line from the reference position to the actual position of the point.
A displacement may be also described as a relative position: the final position of a point ($\vec r_f$) relative to its inBlogger: Membangun Indonesia dengan Fisika - Create postitial position ($\vec r_i$), and a displacement vector can be mathematically defined as the difference between the final and initial position vectors:
$\vec s = \vec r_f - \vec r_i =\Delta \vec r$
In considering motions of objects over time the instantaneous velocity of the object is the rate of change of the displacement as a function of time. The velocity then is distinct from the instantaneous speed which is the time rate of change of the distance traveled along a specific path. The velocity may be equivalently defined as the time rate of change of the position vector. If one considers a moving initial position, or equivalenty a moving origin (e.g. an initial position or origin which is fixed to a train wagon, which in turn moves with respect to its rail track), the velocity of P (e.g. a point representing the position of a passenger walking on the train) may be referred to as a relative velocity, as opposed to an absolute velocity, which is computed with respect to a point which is considered to be "fixed in space" (such as, for instance, a point fixed on the floor of the train station).
For motion over a given interval of time, the displacement divided by the length of the time interval defines the average velocity.

## Distance versus directed distance and displacement

Distance along a path compared with displacement
Distance cannot be negative and distance travelled never decreases. Distance is a scalar quantity or a magnitude, whereas displacement is a vector quantity with both magnitude and direction.
The distance covered by a vehicle (for example as recorded by an odometer), person, animal, or object along a curved path from a point A to a point B should be distinguished from the straight line distance from A to B. For example whatever the distance covered during a round trip from A to B and back to A, the displacement is zero as start and end points coincide. In general the straight line distance does not equal distance travelled, except for journeys in a straight line.

# Velocity

In physics, velocity is speed in a given direction. Speed describes only how fast an object is moving, whereas velocity gives both the speed and direction of the object's motion. To have a constant velocity, an object must have a constant speed and motion in a constant direction. Constant direction, typically constrains the object to motion in a straight path. A car moving at a constant 20 kilometers per hour in a circular path does not have a constant velocity. The rate of change in velocity is acceleration. Velocity is a vector physical quantity; both magnitude and direction are required to define it. The scalar absolute value (magnitude) of velocity is speed, a quantity that is measured in metres per second (m/s or ms−1) when using the SI (metric) system.
For example, "5 metres per second" is a scalar and not a vector, whereas "5 metres per second east" is a vector. The average velocity v of an object moving through a displacement $( \Delta \mathbf{x})$ during a time interval t) is described by the formula:
$\mathbf{\bar{v}} = \frac{\Delta \mathbf{x}}{\Delta t}.$
The rate of change of velocity is acceleration – how an object's speed or direction of travel changes over time, and how it is changing at a particular point in time

The equations that apply to bodies moving linearly (in one dimension) with constant acceleration are often referred to as "SUVAT" equations where the five variables are represented by those letters (s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time); the five letters may be shown in a different order.
The body is considered between two instants in time: one initial point and one current (or final) point. Problems in kinematics may deal with more than two instants, and several applications of the equations are then required. If a is constant, the differential, a dt, may be integrated over an interval from 0 to Δt (Δt = tti), to obtain a linear relationship for velocity. Integration of the velocity yields a quadratic relationship for position at the end of the interval.
 $v = v_i + a \Delta t \,$$s = s_i + v_i\Delta t + \tfrac{1}{2} a(\Delta t)^2 \,$$s = s_i + \tfrac{1}{2} (v + v_i)\Delta t \,$$v^2 = v_i^2 + 2a(s - s_i) \,$ where... $v_i \,$ is the body's initial velocity$s_i \,$ is the body's initial position and its current state is described by: $v \,$, The velocity at the end of the interval$s \,$, the position at the end of the interval (displacement)$\Delta t \,$, the time interval between the initial and current states$a \,$, the constant acceleration, or in the case of bodies moving under the influence of gravity, g.
Note that each of the equations contains four of the five variables. Thus, in this situation it is sufficient to know three out of the five variables to calculate the remaining two.

### Relative velocity

To describe the motion of object A with respect to object B, when we know how each is moving with respect to a reference object O, we can use vector algebra. Choose an origin for reference, and let the positions of objects A, B, and O be denoted by rA, rB, and rO. Then the position of A relative to the reference object O is
$\mathbf{r}_{A/O} = \mathbf{r}_{A} - \mathbf{r}_{O} \,\!$
Consequently, the position of A relative to B is
$\mathbf{r}_{A/B} = \mathbf{r}_A - \mathbf{r}_B = \mathbf{r}_A - \mathbf{r}_O - \left(\mathbf{r}_B-\mathbf{r}_O\right) = \mathbf{r}_{A/O}-\mathbf{r}_{B/O} \ .$
The above relative equation states that the motion of A relative to B is equal to the motion of A relative to O minus the motion of B relative to O. It may be easier to visualize this result if the terms are re-arranged:
$\mathbf{r}_{A/O} = \mathbf{r}_{A/B} + \mathbf{r}_{B/O} \ ,$
or, in words, the motion of A relative to the reference is that of B plus the relative motion of A with respect to B. These relations between displacements become relations between velocities by simple time-differentiation, and a second differentiation makes them apply to accelerations.
For example, let Ann move with velocity $\mathbf{V}_{A}$ relative to the reference (we drop the O subscript for convenience) and let Bob move with velocity $\mathbf{V}_{B}$, each velocity given with respect to the ground (point O). To find how fast Ann is moving relative to Bob (we call this velocity $\mathbf{V}_{A/B}$), the equation above gives:
$\mathbf{V}_{A} = \mathbf{V}_{B} + \mathbf{V}_{A/B} \,\! .$
To find $\mathbf{V}_{A/B}$ we simply rearrange this equation to obtain:
$\mathbf{V}_{A/B} = \mathbf{V}_{A} -\mathbf{V}_{B} \,\! .$
At v

# Motion graphs and derivatives

In mechanics, the derivative of the position vs. time graph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the y-axis and time on the x-axis, the slope of the curve is given by:
$v = \frac{\Delta y}{\Delta x} = \frac{\Delta s}{\Delta t}.$
Here s is the position of the object, and t is the time. Therefore, the slope of the curve gives the change in position (in metres) divided by the change in time (in seconds), which is the definition of the average velocity (in meters per second $(\begin{matrix} \frac{m}{s} \end{matrix})$) for that interval of time on the graph. If this interval is made to be infinitesimally small, such that Δs becomes ds and Δt becomes dt, the result is the instantaneous velocity at time t, or the derivative of the position with respect to time.
A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in y over change in x:
$a = \frac{\Delta y}{\Delta x} = \frac{\Delta v}{\Delta t}.$
Where v is the velocity, measured in $\begin{matrix} \frac{m}{s} \end{matrix}$, and t is the time measured in seconds. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives $\begin{matrix} \frac{dv}{dt} \end{matrix}$, the instantaneous acceleration at time t, or the derivative of the velocity with respect to time (or the second derivative of the position with respect to time). The units of this slope or derivative are in meters per second per second ($\begin{matrix} \frac{m}{s^2} \end{matrix}$, usually termed "meters per second-squared"), and so, therefore, is the acceleration.
Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the seconds cancel out and only meters remain. $\begin{matrix} \frac{m}{s} \end{matrix}s = m$.)
The same multiplication rule holds true for acceleration vs. time graphs. When $(\begin{matrix} \frac{m}{s^2} \end{matrix})$ is multiplied by time (s), velocity is obtained. ($\begin{matrix} \frac{m}{s^2} \end{matrix}s = \begin{matrix} \frac{m}{s} \end{matrix}$).

Average acceleration is the change in velocity (Δv) divided by the change in time (Δt). Instantaneous acceleration is the acceleration at a specific point in time which is for a very short interval of time as Δt approaches zero.

## Tangential and centripetal acceleration

The velocity of a particle moving on a curved path as a function of time can be written as:
$\mathbf{v} (t) =v(t) \frac {\mathbf{v}(t)}{v(t)} = v(t) \mathbf{u}_\mathrm{t}(t) ,$
with v(t) equal to the speed of travel along the path, and
$\mathbf{u}_\mathrm{t} = \frac {\mathbf{v}(t)}{v(t)} \ ,$
a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. Taking into account both the changing speed v(t) and the changing direction of ut, the acceleration of a particle moving on a curved path on a planar surface can be written using the chain rule of differentiation[4] and the derivative of the product of two functions of time as:
\begin{alignat}{3} \mathbf{a} & = \frac{d \mathbf{v}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t} +v(t)\frac{d \mathbf{u}_\mathrm{t}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t}+ \frac{v^2}{R}\mathbf{u}_\mathrm{n}\ , \\ \end{alignat}
where un is the unit (inward) normal vector to the particle's trajectory, and R is its instantaneous radius of curvature based upon the osculating circle at time t. These components are called the tangential acceleration and the radial acceleration or centripetal acceleration (see also circular motion and centripetal force).
Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet-Serret formulas.[5][6]

## Variable rates of change

In this example, the yellow area represents the displacement of the object as it moves. (The distance can be measured by taking the absolute value of the function.) The three green lines represent the values for acceleration at different points along the curve.
The expressions given above apply only when the rate of change is constant or when only the average (mean) rate of change is required. If the velocity or positions change non-linearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. Differentiation reduces the time-spans used above to be extremely small and gives a velocity or acceleration at each point on the graph rather than between a start and end point. The derivative forms of the above equations are
$v = \frac{ds}{dt},$
$a = \frac{dv}{dt}.$
Since acceleration differentiates the expression involving position, it can be rewritten as a second derivative with respect to position:
Blogger: Membangun Indonesia dengan Fisika - Create post$a = \frac{d^2 s}{dt^2}.$
Since, for the purposes of mechanics such as this, integration is the opposite of differentiation, it is also possible to express position as a function of velocity and velocity as a function of acceleration. The process of determining the area under the curve, as described above, can give the displacement and change in velocity over particular time intervals by using definite integrals:
$s(t_2)-s(t_1) = \int_{t_1}^{t_2}{v}\, dt,$
$v(t_2)-v(t_1) = \int_{t_1}^{t_2}{a}\, dt.$

## Ideal projectile motion

Ideal projectile motion assumes that there is no air resistance. This assumption simplifies the math greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are small. Ideal projectile motion is also a good introduction to the topic before adding the complications of air resistance.

### Derivations

#### Flat Ground

Range of a projectile (in vacuum).
First we examine the case where (y0) is zero. The horizontal position 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 factoring:
$\frac{} {}t = 0$
or
$t = \frac{2 v \sin \theta} {g}$
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(x + y) = sin(x)cos(y) + sin(y)cos(x)

If x and y are same,
sin(2x) = 2sin(x)cos(x)
allows us to simplify the solution to
$d = \frac {v^2} {g} \sin \left(2 \theta\right)$
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
$t = \frac {v \sin \theta} {g} + \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}$
Solving for the range once again
$d = \frac {v \cos \theta} {g} \left [ v \sin \theta + \sqrt{\left(v \sin \theta \right)^2 + 2 g y_0} \right]$

#### Maximum Range

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).
For cases where the projectile lands at the same height from which it is launched, the maximum range is obtained by using a launch angle of 45 degrees. A projectile that is launched with an elevation of 0 degrees will strike the ground immediately (range = 0), though it may then bounce or roll. A projectile that is fired with an elevation of 90 degrees (i.e. straight up) will travel straight up, then straight down, and strike the ground at the point from which it is launched, again yielding a range of 0.
The elevation angle which will provide the maximum range when launching the projectile from a non-zero initial height can be computed by finding the derivative of the range with respect to the elevation angle and setting the derivative to zero to find the extremum:

$\frac { dR } { d\theta} = \frac {v^2} {g} [ \cos \theta ( \cos \theta + \frac {\sin \theta \cdot \cos \theta} {\sqrt {(\sin \theta)^2 + C} } ) - \sin \theta ( \sin \theta + \sqrt { ( \sin \theta )^2 + C })]$
where $C = \frac {2 g y_0} { v^2}$ and $\,\!R =$ horizontal range.
Setting the derivative to zero provides the equation:

$(\cos \theta)^2 + \frac {\sin \theta \cdot (\cos \theta )^2} {\sqrt {(\sin \theta )^2 + C} } - ( \sin \theta )^2 - \sin \theta \sqrt {(\sin \theta )^2 + C} = 0$

Substituting u = (cos θ)2 and 1 − u = (sin θ)2 produces:
$u + \frac {u \sqrt {1-u}} {\sqrt {1-u+C}} - (1-u) - (\sqrt {1-u}) \sqrt {1-u+C} = 0$
Which reduces to the extremely surprisingly simple expression:
$u = \frac {C + 1} {C + 2}$
Replacing our substitutions yields the angle that produces the maximum range for uneven ground, ignoring air resistance:
$\theta = \arccos \sqrt { \frac {2 g y_0 + v^2} {2 g y_0 + 2 v^2} }$
Note that for zero initial height, the elevation angle that produces maximum range is 45 degrees, as expected. For positive initial heights, the elevation angle is below 45 degrees, and for negative initial heights (bounded below by y0 > − 0.5v2 / g), the elevation angle is greater than 45 degrees.
Example: For the values g = 9.81m / s2, y0 = 40m , and v = 50m / s, an elevation angle θ = 41.1° produces a maximum range of Rmax = 292.1m.

#### Angle of impact

The angle ψ at which the projectile lands is given by:
$\tan \psi = \frac {-v_y(t_d)} {v_x(t_d)} = \frac {\sqrt { v^2 (\sin \theta)^2 + 2 g y_0 }} { v \cos \theta}$
For maximum range, this results in the following equation:
$(\tan \psi)^2 = \frac { 2 g y_0 + v^2 } { v^2 } = C+1$
Rewriting the original solution for θ, we get:
$(\tan \theta)^2 = \frac { 1 - (\cos \theta)^2 } { (\cos \theta)^2 } = \frac { v^2 } { 2 g y_0 + v^2 } = \frac { 1 } { C + 1 }$
Multiplying with the equation for (tan ψ)^2 gives:
$(\tan \psi)^2 (\tan \theta)^2 = \frac { 2 g y_0 + v^2 } { v^2 } \frac { v^2 } { 2 g y_0 + v^2 } = 1$
Because of the trigonometric identity
$\tan (\theta + \psi) = \frac { \tan \theta + \tan \psi } { 1 - \tan \theta \tan \psi }$,
this means that θ + ψ must be 90 degrees.

# Trajectory of a projectile

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).
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.[1]
The following applies for ranges which are small compared to the size of the Earth.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

## Conditions at the final position of the projectile

### [Distance travelled

The total horizontal distance (d) travelled.
$d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)$
When the surface the object is launched from and is flying over is flat (the initial height is zero), the distance travelled 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 (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
vf = vi + 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 θ required to hit coordinate (x,y)

Vacuum trajectory of a projectile for different launch angles
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:
$\theta = \arctan{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)}$
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, 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 θ
$0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y$ (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 = rcos ϕ and y = rsin ϕ and substitute to get:
$\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi}\right)}$

### Uphill/downhill in uniform gravity in a vacuum

Illustration showing the trajectory of a bullet fired at an uphill target

Given a hill angle α and launch angle θ as before, it can be shown that the range along the hill Rs forms a ratio with the original range R along the imaginary horizontal, such that:
$\frac{R_s} {R}=(1-\cot \theta \tan \alpha)\sec \alpha$ (Equation 11)
In this equation, downhill occurs when α is between 0 and -90 degrees. For this range of α we know: tan( − α) = − tan α and sec( − α) = sec α. Thus for this range of α, Rs / R = (1 + tan θtan α)sec α. Thus Rs / R is a positive value meaning the range downhill is always further than along level terrain. The lower level of terrain causes the projectile to remain in the air longer, allowing it to travel further horizontally before hitting the ground.
While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to Rs / R = 1 (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" θcr:
$1=(1-\tan \theta \tan \alpha)\sec \alpha \quad \;$
$\theta_{cr}=\arctan((1-\csc \alpha)\cot \alpha) \quad \;$
Equation 11 may also be used to develop the "rifleman's rule" for small values of α and θ (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both tan α and tanθ have a small value and thus when multiplied together (as in equation 11), the result is almost zero. Thus equation 11 may be approximated as:
$\frac{R_s} {R}=(1-0)\sec \alpha$
And solving for level terrain range, R
$R=R_s \cos \alpha \$ "Rifleman's rule"
Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position."[1]

#### Derivation based on equations of a parabola

The intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in Cartesian coordinates (Equation 10) intersecting the hill of slope m in standard linear form at coordinates (x,y):
$y=mx+b \;$ (Equation 12) where in this case, y = dv, x = dh and b = 0
Substituting the value of dv = mdh into Equation 10:
$m x=-\frac{g}{2v^2{\cos}^2 \theta}x^2 + \frac{\sin \theta}{\cos \theta} x$
$x=\frac{2v^2\cos^2\theta}{g}\left(\frac{\sin \theta}{\cos \theta}-m\right)$ (Solving above x)
This value of x may be substituted back into the linear equation 12 to get the corresponding y coordinate at the intercept:
$y=mx=m \frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right)$
Now the slant range Rs is the distance of the intercept from the origin, which is just the hypotenuse of x and y:
$R_s=\sqrt{x^2+y^2}=\sqrt{\left(\frac{2v^2\cos^2\theta}{g}\left(\frac{\sin \theta}{\cos \theta}-m\right)\right)^2+\left(m \frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right)\right)^2}$
$=\frac{2v^2\cos^2\theta}{g} \sqrt{\left(\frac{\sin \theta}{\cos \theta}-m\right)^2+m^2 \left(\frac{\sin \theta}{\cos \theta}-m\right)^2}$
$=\frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right) \sqrt{1+m^2}$
Now α is defined as the angle of the hill, so by definition of tangent, m = tan α. This can be substituted into the equation for Rs:
$R_s=\frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-\tan \alpha\right) \sqrt{1+\tan^2 \alpha}$
Now this can be refactored and the trigonometric identity for $\sec \alpha = \sqrt {1 + \tan^2 \alpha}$ may be used:
$R_s=\frac{2v^2\cos\theta\sin\theta}{g}\left(1-\frac{\sin\theta}{\cos\theta}\tan\alpha\right)\sec\alpha$
Now the flat range R = v2sin 2θ / g = 2v2sin θcos θ / g by the previously used trigonometric identity and cos θ / sin θ = cotanθ so:
$R_s=R(1-\cot\theta\tan\alpha)\sec\alpha \;$
$\frac{R_s}{R}=(1-\cot\theta\tan\alpha)\sec\alpha$

.

## Trajectory of a projectile with air resistance

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, v0,vx and vy 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. Fair = − kv 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 ΣF = − kvx = max and in the y direction ΣF = − kvymg = may.

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 vx and, subsequently, x will not be enumerated. Given the initial conditions vx = vxo (where vxo is understood to be the x component of the initial velocity) and sx = 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 vy = vyo and sy = 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 red 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 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 p2. In the first example of proportionality, where the velocity was doubled 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.