Editing Projectile motion (section)

=== 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>