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Angle of elevation of Trajectory

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Notes:

In terms of angle of elevation $θ$ and initial speed $v$ :

$v_h=v \cos \theta,\quad v_v=v \sin \theta \;$

giving the range as

$R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.$

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

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

Note that the sine function is such that there are two solutions for $θ$ for a given range $d h$ . Physically, this corresponds to a direct shot versus a mortar shot up and over obstacles to the target. The angle $θ$ giving the maximum range can be found by considering the derivative or $R$ with respect to $θ$ and setting it to zero.

${\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0$

which has a non trivial solutions at $2\theta=\pi/2=90^\circ$ . The maximum range is then $R_{max} = v^2/g\,$ . At this angle $s i n (π / 2) = 1$ so the maximum height obtained is ${v^2 \over 4g}$ .

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height $H = v s i n (θ) / (2 g)$ with respect to $θ$ , that is ${\mathrm{d}H\over \mathrm{d}\theta}=v \cos(\theta) /(2g)$ which is zero when $\theta=\pi=180^\circ$ . So the maximum height $H_{max}={v\over 2g}$
is obtain when the projectile is fired straight up. The equation of the
trajectory of a projectile fired in uniform gravity in a vacuum on
Earth in Cartesian coordinates is