Editing Projectile motion (section)

=== 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).