[see the attached figure for a better understanding of my notations]
All the angles below are in radians unless written otherwise
[tex]m[/tex] : mass of the ball
[tex]\alpha[/tex] : angle by which the ground is inclined (here [tex]a=-35^\circ[/tex])
[tex]\beta[/tex] : initial angle of the throw
[tex]g[/tex] : acceleration due to gravity, [tex]g\approx9.81m^2/s[/tex]
[tex]\vec{y}[/tex] : vertical unit vector, pointing upwards
[tex]\vec{x}[/tex] : horizontal unit vector, pointing to the right
[tex]x,y[/tex] : x,y-position of the ball
[tex]\vec{a}[/tex] : ball's acceleration
[tex]\vec{v}[/tex] : ball's speed
[tex]\vec{p}[/tex] : ball's position
[tex]\vec{v_0}[/tex] : initial ball's speed
I don't now how much you know in physics, so I'll try to break this down as much as possible.
The only force that is applied on the ball when it's falling is gravity [tex]\vec{P}[/tex]. By Newton's second law, [tex]m\vec{a}=\vec{P}=-mg\vec{y}[/tex] hence by integrating with respect to time, [tex]\vec{v}=-gt\vec{y}+\vec{v_0}[/tex] which by integrating once more yields [tex]\vec{p}=-g\frac{t^2}2\vec{y}+\vec{v_0}t[/tex]
Thus [tex]y=-g\frac{t^2}2+v_0\sin(\beta)t,x=v_0\cos(\beta)t[/tex]. That last one gives [tex]t=\dfrac{x}{v_0\cos(\beta)}[/tex] hence [tex]y=-g\dfrac{x^2}{2v_0^2\cos^2(\beta)}+x\tan(\beta)[/tex]
The ball stops when it intersects with the slope defined by [tex]y=\tan(\alpha)x[/tex] hence if we call [tex]X[/tex] the abscissa of the the intersection point, [tex]\tan(\alpha)X=-g\dfrac{X^2}{2v_0^2\cos^2(\beta)}+X\tan(\beta)[/tex] hence [tex]X(\beta)=\dfrac{(b-a)2v_0^2\cos^2(\beta)}{g}[/tex] where [tex]a=\tan(\alpha),b=\tan(\beta)[/tex].
To find the maximum of [tex]X(\beta)[/tex], we derive the function and solve [tex]\dfrac{dX}{d\beta}=0[/tex] which is equivalent to [tex]0=\dfrac{d((\tan(\beta)-a)\cos^2(\beta))}{d\beta}=1-(b-a)\sin(2\beta)[/tex]. Using the well known formula [tex]\sin(2\beta)=\dfrac{2b}{1+b^2}[/tex] we get [tex]b^2-2ab-1=0[/tex]. The discriminant of this equation is [tex]\Delta=4(a^2+1)[/tex] hence the unique positive solution is [tex]b=a+\sqrt{a^2+1}[/tex] thus [tex]\beta=\arctan(a+\sqrt{a^2+1})[/tex]
In our case [tex]a=\tan(\dfrac{-35*2\pi}{360})=-\tan(\dfrac{7\pi}{36})[/tex] hence [tex]b=\arctan\left(a+\sqrt{a^2+1}\right)=27.5^\circ[/tex], which is the optimum angle to get a maximum range.
