I'm trying to determine a general equation for projectile motion with the angle $\theta$ as the independent variable/unknown. I'm having trouble trying to isolate $\theta$, here is my understanding so far.
First, derive the equations of motion for the horizontal component of the projectile:
$$\frac{d^2x}{dt^2}=0$$
$$\frac{dx}{dt}=v*\cos(\theta)$$
$$x=vt*\cos(\theta)+x_i$$
Where $x_i$ is the initial horizontal distance which is assumed to be 0, thus the horizontal position can be rewritten as a function of $t$:
$$Eq.(1):\qquad t=\frac{x}{v*\cos(\theta)}$$
Vertical components of motion for the projectile:
$$\frac{d^2y}{dt^2}=-g$$
$$\frac{dy}{dt}=-gt+v*\sin(\theta)$$
$$y=-0.5gt^2+vt*\sin(\theta)+y_i$$
Where $y_i$ is the initial vertical distance which is assumed to be 0. Thus:
$$Eq.(2)\qquad y=-0.5gt^2+vt*\sin(\theta)$$
Substituting Eq.(1) into Eq.(2):
$$y=-0.5g\left( \frac{x}{v*\cos(\theta)}\right)^2+\frac{x*\sin(\theta)}{\cos(\theta)}$$
…and the trig can be simplified down to functions of tan:
$$y=\frac{-0.5gx^2(1+\tan(\theta)^2)}{v^2}+x*\tan(\theta)$$
$$y=\frac{-0.5gx^2}{v^2}+\frac{-0.5gx^2*\tan(\theta)^2}{v^2}+x*\tan(\theta)$$
$$0=\frac{-0.5gx^2}{v^2}\tan(\theta)^2+x*\tan(\theta)+\frac{-0.5gx^2}{v^2}-y$$
$g$ is the acceleration due to gravity.
$x$ is the horizontal distance traveled.
$y$ is the vertical distance traveled.
$v$ is the initial velocity launched at angle $\theta$ (note that we assume final velocity in both directions is 0).
$\theta$ is the launch angle.
So, I end up with the above – a nasty looking quadratic. Again, this general equation assumes that all other variables except angle $\theta$ are known.
My question is, am I on the right track or have I made a mistake as the algebra starts getting a little annoying with the next step.
Could I now use the quadratic formula to isolate $\theta$ with the quadratic's respective coefficients given as $a=\frac{-0.5gx^2}{v^2}$, $b=x$ and $c=\frac{-0.5gx^2}{v^2}-y$?
Please help me, I am melting 🙁
Best Answer
Using the relation,$$y=x\tan \theta -\frac{gx^2}{ 2u^2 \cos^2\theta}$$, $$\implies y=x\tan\theta-\frac{gx^2}{2u^2}(1+\tan^2\theta)$$ Let $k=\frac{gx^2}{2u^2}$ and our quadratic then becomes, $$y=x\tan\theta-k-k\tan^2\theta$$ $$\implies k\tan^2\theta-x\tan\theta+(k+y)=0$$ Which gives, $$\tan \theta =\frac{x \pm \sqrt{x^2-4ky-4k^2}}{2k}$$ and you can put back $k=\frac{gx^2}{2u^2}$