Editing Trajectory (section)

====Angle of elevation====
[[File:Selomie Melkie - Forensics Final Project (5).jpg|thumb|An example showing how to calculate bullet trajectory]]
In terms of angle of elevation <math>\theta</math> and initial speed <math>v</math>:

:<math>v_h=v \cos \theta,\quad v_v=v \sin \theta \;</math>

giving the range as

:<math>R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.</math>

This equation can be rearranged to find the angle for a required range

: <math> \theta =  \frac 1 2 \sin^{-1} \left( \frac{g R}{ v^2 } \right) </math> (Equation II: angle of projectile launch)

Note that the [[sine]] function is such that there are two solutions for <math>\theta</math> for a given range <math>d_h</math>.  The angle <math>\theta</math> giving the maximum range can be found by considering the derivative or <math>R</math> with respect to <math>\theta</math> and setting it to zero.

:<math>{\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0</math>

which has a nontrivial solution at <math>2\theta=\pi/2=90^\circ</math>, or <math>\theta=45^\circ</math>. The maximum range is then <math>R_{\max} = v^2/g\,</math>. At this angle <math>\sin(\pi/2)=1</math>, so the maximum height obtained is <math>{v^2 \over 4g}</math>.

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height <math>H=v^2 \sin^2(\theta) /(2g)</math> with respect to <math>\theta</math>, that is
<math>{\mathrm{d}H\over \mathrm{d}\theta}=v^2 2\cos(\theta)\sin(\theta) /(2g)</math>
which is zero when <math>\theta=\pi/2=90^\circ</math>. So the maximum height <math>H_\mathrm{max}={v^2\over 2g}</math> is obtained when the projectile is fired straight up.