Editing Projectile motion (section)

== Properties of the trajectory ==
=== Time of flight or total time of the whole journey ===
The total time <var>t</var> for which the projectile remains in the air is called the time-of-flight.
:<math> y=v_0 t\sin(\theta) -\frac{1}{2} g t^2 </math>

After the flight, the projectile returns to the horizontal axis (x-axis), so <math> y=0 </math>.
:<math> v_0 t\sin(\theta) =\frac{1}{2} g t^2 </math>

:<math> v_0 \sin(\theta) = \frac{1}{2}gt </math>

:<math> t=\frac{2 v_0\sin(\theta)}{g} </math>

Note that we have neglected air resistance on the projectile.

If the starting point is at height <var>y<sub>0</sub></var> with respect to the point of impact, the time of flight is:
: <math> t = \frac{d}{v \cos\theta} = \frac{v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0}}{g} </math>

As above, this expression can be reduced (<var>y<sub>0</sub></var> is 0) to

: <math> t = \frac{v\sin{\theta} + \sqrt{(v\sin{\theta})^2}}{g} = \frac{2v\sin{\theta}}{g}, </math> = <math> \frac{2v\sin{(45^\circ)}}{|g|} = \frac{\sqrt{2}v}{|g|}</math>

if <var>θ</var> equals 45°.

=== Time of flight to the target's position ===
As shown above in the '''Displacement''' section, the horizontal and vertical velocity of a projectile are independent of each other.

Because of this, we can find the time to reach a target using the displacement formula for the horizontal velocity:

<math>x = v_0 t \cos(\theta)</math>

<math>\frac{x}{t}=v_0\cos(\theta)</math>

<math>t=\frac{x}{v_0\cos(\theta)}</math>

This equation will give the total time ''t'' the projectile must travel for to reach the target's horizontal displacement, neglecting air resistance.

=== Maximum height of projectile ===
[[File:Ferde hajitas4.svg|thumb|250px|Maximum height of projectile]]
The greatest height that the object will reach is known as the peak of the object's motion.
The increase in height will last until <math> v_y=0 </math>, that is,
:<math> 0=v_0 \sin(\theta) - gt_h </math>.
Time to reach the maximum height(h):
:<math display="inline"> t_h = \frac{v_0 \sin(\theta)}{|g|} </math>.
For the vertical displacement of the maximum height of the projectile:
:<math> h = v_0 t_h \sin(\theta) - \frac{1}{2} gt^2_h </math>
:<math> h = \frac{v_0^2 \sin^2(\theta)}{2|g|} </math>
The maximum reachable height is obtained for ''&theta;''=90°:
:<math> h_{\mathrm{max}} = \frac{v_0^2}{2|g|} </math>

If the projectile's position (x,y) and launch angle (θ) are known, the maximum height can be found by solving for h in the following equation:
:<math>h=\frac{(x\tan\theta)^2}{4(x\tan\theta-y)}. </math>

Angle of [[elevation]] (φ) at the maximum height is given by:
:<math>\phi = \arctan{{\tan\theta\over 2}}</math>
{{clear}}

=== Relation between horizontal range and maximum height ===
The relation between the range <var>d</var> on the horizontal plane and the maximum height <var>h</var> reached at <math> \frac{t_d}{2} </math> is:
:<math> h = \frac{d\tan\theta}{4} </math>

{{Collapse top|title=Proof}}
<math> h = \frac{v_0^2\sin^2\theta}{2|g|} </math>
:<math> d = \frac{v_0^2\sin2\theta}{|g|} </math>
:<math> \frac{h}{d} = \frac{v_0^2\sin^2\theta}{2|g|} </math> × <math> \frac{g}{v_0^2\sin2\theta} </math>
:<math> \frac{h}{d} = \frac{\sin^2\theta}{4\sin\theta\cos\theta} </math>
<math> h = \frac{d\tan\theta}{4} </math>.

If <math>h = R</math>
:'''<math> \theta = \arctan(4)\approx 76.0^\circ </math>'''
{{Collapse bottom|}}

=== Maximum distance of projectile ===
{{Main|Range of a projectile}}

[[File:Ferde hajitas5.svg|thumb|250px|The maximum distance of projectile]]

The range and the maximum height of the projectile do not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction. The horizontal range <var>d</var> of the projectile is the horizontal distance it has traveled when it returns to its initial height (<math display="inline">y=0</math>).
: <math> 0 = v_0 t_d \sin(\theta) - \frac{1}{2}gt_d^2 </math>.

Time to reach ground:
: <math> t_d = \frac{2v_0 \sin(\theta)}{|g|} </math>.
From the horizontal displacement the maximum distance of the projectile:
: <math> d = v_0 t_d \cos(\theta) </math>,
so{{NoteTag|<math> 2\cdot\sin(\alpha)\cdot\cos(\alpha) = \sin(2\alpha) </math>}}
: <math> d = \frac{v_0^2}{|g|}\sin(2\theta). </math>

Note that <var>d</var> has its maximum value when
: <math> \sin(2\theta)=1, </math>
which necessarily corresponds to <math display="inline"> 2\theta=90^\circ </math>, or <math display="inline"> \theta=45^\circ </math>.[[File:Ideal projectile motion for different angles.svg|thumb|350px|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/s<sup>2</sup>. 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 <var>(d)</var> traveled.

: <math> d = \frac{v \cos \theta}{|g|} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right) </math>

When the surface is flat (initial height of the object is zero), the distance traveled:<ref>{{cite book |last=Tatum |url=https://orca.phys.uvic.ca/~tatum/classmechs/class7.pdf |title=Classical Mechanics |year=2019 |pages=Ch. 7}}</ref>

: <math> d = \frac{v^2 \sin(2 \theta)}{|g|} </math>

Thus the maximum distance is obtained if <var>θ</var> is 45 degrees. This distance is:

: <math> d_{\mathrm{max}} = \frac{v^2}{|g|} </math>

=== Application of the work energy theorem ===
According to the [[Work (physics)#Work–energy principle|work-energy theorem]] the vertical component of velocity is:
:<math> v_y^2 = (v_0 \sin \theta)^2-2gy </math>.

These formulae ignore [[Drag (physics)|aerodynamic drag]] and also assume that the landing area is at uniform height 0.

=== Angle of reach ===
The "angle of reach" is the angle (<var>θ</var>) at which a projectile must be launched in order to go a distance <var>d</var>, given the initial velocity <var>v</var>.

: <math> \sin(2\theta) = \frac{gd}{v^2} </math>
There are two solutions:
: <math> \theta = \frac{1}{2} \arcsin \left( \frac{gd}{v^2} \right) </math> (shallow trajectory)
and because <math> \sin(2\theta) = \cos (2\theta - 90^\circ )</math>,

: <math> \theta = 45^\circ + \frac{1}{2} \arccos \left( \frac{gd}{v^2} \right) </math> (steep trajectory)

=== Angle <var>θ</var> required to hit coordinate (<var>x</var>, <var>y</var>) ===
[[Image:Trajectory for changing launch angle.gif|right|thumb|330px|Vacuum trajectory of a projectile for different launch angles. Launch speed is the same for all angles, 50&nbsp;m/s, and "g" is 10&nbsp;m/s<sup>2</sup>.]]
To hit a target at range <var>x</var> and altitude <var>y</var> when fired from (0,0) and with initial speed <var>v,</var> the required angle(s) of launch <var>θ</var> are:

: <math> \theta = \arctan{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)} </math>

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 (<var>x</var>,<var>y</var>) selected. This formula allows one to find the angle of launch needed without the restriction of <math display="inline"> y= 0 </math>.

One can also ask what launch angle allows the lowest possible launch velocity. This occurs when the two solutions above are equal, implying that the quantity under the square root sign is zero. This, tan θ = v<sup>2</sup>/gx, requires solving a quadratic equation for <math> v^2 </math>,<ref>with V=v<sup>2</sup>, V<sup>2</sup> - 2gy.V - (gx)<sup>2</sup> = 0: <math>V_{1,2}=(2gy \pm \sqrt{4g^2y^2 + 4(gx)^2})/2</math></ref> and we find
:<sup><math> v^2/g = (2gy\pm g\sqrt{4y^2+4(1x)^2})/2g = y\pm\sqrt{y^2+x^2}. </math></sup>
This gives
:<math> \theta=\arctan\left(y/x+\sqrt{y^2/x^2+1}\right). </math>

If we denote the angle whose tangent is {{mvar|y/x}} by {{mvar|α}},<ref><math display="inline">tan(\theta) = y/x+\sqrt{(y/x)^2+1} </math> where <math display="inline">y/x = sin(\alpha)/cos(\alpha): sin^2\alpha+cos^2\alpha=1=(tan^2\alpha+1)cos^2\alpha</math></ref> then
:<math> \tan\theta=\frac{\sin\alpha+1}{\cos\alpha}, </math> its reciprocal:
:<math> \tan(\pi/2-\theta)=\frac{\cos\alpha}{\sin\alpha+1}, </math><ref><math>1/\tan(\pi/2-\theta)=\cos(\pi/2-\theta)/\sin(\pi/2-\theta)=\sin\theta/\cos\theta</math>

<math>
=(\sin\alpha+1)/\cos\alpha=\tan\alpha+1/\cos\alpha</math>

<math>=\tan\theta</math></ref>
:<math> \cos^2(\pi/2-\theta)=\frac 12(\sin\alpha+1) </math>
:<math> 2\cos^2(\pi/2-\theta)-1=\cos(\pi/2-\alpha) </math>
This implies
:<math> \theta = \pi/2 - \frac 12(\pi/2-\alpha). </math>

In other words, the launch should be at the angle <math display="inline"> (\pi/2 + \alpha)/2 </math> halfway between the target and [[zenith]] (vector opposite to gravity).

=== Total Path Length of the Trajectory ===
The length of the parabolic arc traced by a projectile, <var>L</var>, given that the height of launch and landing is the same (there is no air resistance), is given by the formula:

:<math>L = \frac{v_0^2}{2g} \left( 2\sin\theta + \cos^2\theta\cdot\ln \frac{1 + \sin\theta}{1 - \sin\theta} \right) = \frac{v_0^2}{g} \left( \sin\theta + \cos^2\theta\cdot\tanh^{-1}(\sin\theta) \right)</math>

where <math>v_0</math> is the initial velocity, <math>\theta</math> is the launch angle and <math>g</math> is the acceleration due to gravity as a positive value. The expression can be obtained by evaluating the [[Arc length|arc length integral]] for the height-distance parabola between the bounds ''initial'' and ''final'' displacement (i.e. between 0 and the horizontal range of the projectile) such that:

:<math>L = \int_{0}^{\mathrm{range}} \sqrt{1 + \left ( \frac{\mathrm{d}y}{\mathrm{d}x} \right )^2}\,\mathrm{d}x = \int_{0}^{v_0^2 \sin(2\theta)/g} \sqrt{1+\left (\tan\theta -{g\over {v_0^2 \cos^2\theta}}x\right)^2}\,\mathrm{d}x .</math>

If the time-of-flight is ''t'',
:<math>L = \int_{0}^{t} \sqrt{v_x^2 + v_y^2}\,\mathrm{d}t = \int_{0}^{2v_0\sin\theta/g} \sqrt{(gt)^2-2gv_0\sin\theta t+v_0^2}\,\mathrm{d}t.</math>