Hello, and welcome to FMH!
Let's let
θ=A⋅180π and write:
[MATH]y=h-4.905\left(\frac{x}{v}\right)^2\sec^2(\theta)+\tan(\theta)x[/MATH]
Using a Pythagorean identity, we may write:
[MATH]y=h-4.905\left(\frac{x}{v}\right)^2(\tan^2(\theta)+1)+\tan(\theta)x[/MATH]
Arrange in standard form as quadratic in
tan(θ):
[MATH]4.905\left(\frac{x}{v}\right)^2\tan^2(\theta)-x\tan(\theta)+4.905\left(\frac{x}{v}\right)^2+y-h=0[/MATH]
Now, apply the quadratic formula to get the two roots:
[MATH]\tan(\theta)=\frac{x\pm\sqrt{x^2-4\left(4.905\left(\frac{x}{v}\right)^2\right)\left(4.905\left(\frac{x}{v}\right)^2+y-h\right)}}{2\left(4.905\left(\frac{x}{v}\right)^2\right)}[/MATH]
Hence:
[MATH]\theta=\arctan\left(\frac{x\pm\sqrt{x^2-4\left(4.905\left(\frac{x}{v}\right)^2\right)\left(4.905\left(\frac{x}{v}\right)^2+y-h\right)}}{2\left(4.905\left(\frac{x}{v}\right)^2\right)}\right)[/MATH]
[MATH]A=\frac{180}{\pi}\arctan\left(\frac{x\pm\sqrt{x^2-4\left(4.905\left(\frac{x}{v}\right)^2\right)\left(4.905\left(\frac{x}{v}\right)^2+y-h\right)}}{2\left(4.905\left(\frac{x}{v}\right)^2\right)}\right)[/MATH]