<em>[see the attached figure for a better understanding of my notations]
</em><span>
All the angles below are in radians unless written otherwise
</span><span>
![m](https://tex.z-dn.net/?f=m)
: mass of the ball
</span><span>
![\alpha](https://tex.z-dn.net/?f=%5Calpha)
: angle by which the ground is inclined (here
![a=-35^\circ](https://tex.z-dn.net/?f=a%3D-35%5E%5Ccirc)
)
</span><span>
![\beta](https://tex.z-dn.net/?f=%5Cbeta)
: initial angle of the throw
</span><span>
![g](https://tex.z-dn.net/?f=g)
: acceleration due to gravity,
![g\approx9.81m^2/s](https://tex.z-dn.net/?f=g%5Capprox9.81m%5E2%2Fs)
</span><span>
![\vec{y}](https://tex.z-dn.net/?f=%5Cvec%7By%7D)
: vertical unit vector, pointing upwards
</span><span>
![\vec{x}](https://tex.z-dn.net/?f=%5Cvec%7Bx%7D)
: horizontal unit vector, pointing to the right
</span><span>
![x,y](https://tex.z-dn.net/?f=x%2Cy)
: x,y-position of the ball
</span><span>
![\vec{a}](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D)
: ball's acceleration
</span><span>
![\vec{v}](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D)
: ball's speed
</span><span>
![\vec{p}](https://tex.z-dn.net/?f=%5Cvec%7Bp%7D)
: ball's position
</span><span>
![\vec{v_0}](https://tex.z-dn.net/?f=%5Cvec%7Bv_0%7D)
: initial ball's speed
</span><span>
I don't now how much you know in physics, so I'll try to break this down as much as possible.
</span><span>
The only force that is applied on the ball when it's falling is gravity
![\vec{P}](https://tex.z-dn.net/?f=%5Cvec%7BP%7D)
.</span><span> By Newton's second law,
![m\vec{a}=\vec{P}=-mg\vec{y}](https://tex.z-dn.net/?f=m%5Cvec%7Ba%7D%3D%5Cvec%7BP%7D%3D-mg%5Cvec%7By%7D)
hence by integrating with respect to time,
![\vec{v}=-gt\vec{y}+\vec{v_0}](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D%3D-gt%5Cvec%7By%7D%2B%5Cvec%7Bv_0%7D)
which by integrating once more yields
![\vec{p}=-g\frac{t^2}2\vec{y}+\vec{v_0}t](https://tex.z-dn.net/?f=%5Cvec%7Bp%7D%3D-g%5Cfrac%7Bt%5E2%7D2%5Cvec%7By%7D%2B%5Cvec%7Bv_0%7Dt)
</span><span>
Thus
![y=-g\frac{t^2}2+v_0\sin(\beta)t,x=v_0\cos(\beta)t](https://tex.z-dn.net/?f=y%3D-g%5Cfrac%7Bt%5E2%7D2%2Bv_0%5Csin%28%5Cbeta%29t%2Cx%3Dv_0%5Ccos%28%5Cbeta%29t)
. That last one gives
![t=\dfrac{x}{v_0\cos(\beta)}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bx%7D%7Bv_0%5Ccos%28%5Cbeta%29%7D)
hence
![y=-g\dfrac{x^2}{2v_0^2\cos^2(\beta)}+x\tan(\beta)](https://tex.z-dn.net/?f=y%3D-g%5Cdfrac%7Bx%5E2%7D%7B2v_0%5E2%5Ccos%5E2%28%5Cbeta%29%7D%2Bx%5Ctan%28%5Cbeta%29)
</span><span>
The ball stops when it intersects with the slope defined by
![y=\tan(\alpha)x](https://tex.z-dn.net/?f=y%3D%5Ctan%28%5Calpha%29x)
hence if we call
![X](https://tex.z-dn.net/?f=X)
the abscissa of the the intersection point,
![\tan(\alpha)X=-g\dfrac{X^2}{2v_0^2\cos^2(\beta)}+X\tan(\beta)](https://tex.z-dn.net/?f=%5Ctan%28%5Calpha%29X%3D-g%5Cdfrac%7BX%5E2%7D%7B2v_0%5E2%5Ccos%5E2%28%5Cbeta%29%7D%2BX%5Ctan%28%5Cbeta%29)
hence
![X(\beta)=\dfrac{(b-a)2v_0^2\cos^2(\beta)}{g}](https://tex.z-dn.net/?f=X%28%5Cbeta%29%3D%5Cdfrac%7B%28b-a%292v_0%5E2%5Ccos%5E2%28%5Cbeta%29%7D%7Bg%7D)
where
![a=\tan(\alpha),b=\tan(\beta)](https://tex.z-dn.net/?f=a%3D%5Ctan%28%5Calpha%29%2Cb%3D%5Ctan%28%5Cbeta%29)
.
</span><span>
To find the maximum of
![X(\beta)](https://tex.z-dn.net/?f=X%28%5Cbeta%29)
, we derive the function and solve
![\dfrac{dX}{d\beta}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdX%7D%7Bd%5Cbeta%7D%3D0)
which is equivalent to
![0=\dfrac{d((\tan(\beta)-a)\cos^2(\beta))}{d\beta}=1-(b-a)\sin(2\beta)](https://tex.z-dn.net/?f=0%3D%5Cdfrac%7Bd%28%28%5Ctan%28%5Cbeta%29-a%29%5Ccos%5E2%28%5Cbeta%29%29%7D%7Bd%5Cbeta%7D%3D1-%28b-a%29%5Csin%282%5Cbeta%29)
. Using the well known formula
![\sin(2\beta)=\dfrac{2b}{1+b^2}](https://tex.z-dn.net/?f=%5Csin%282%5Cbeta%29%3D%5Cdfrac%7B2b%7D%7B1%2Bb%5E2%7D)
we get
![b^2-2ab-1=0](https://tex.z-dn.net/?f=b%5E2-2ab-1%3D0)
. The discriminant of this equation is
![\Delta=4(a^2+1)](https://tex.z-dn.net/?f=%5CDelta%3D4%28a%5E2%2B1%29)
hence the unique positive solution is
![b=a+\sqrt{a^2+1}](https://tex.z-dn.net/?f=b%3Da%2B%5Csqrt%7Ba%5E2%2B1%7D)
thus
![\beta=\arctan(a+\sqrt{a^2+1})](https://tex.z-dn.net/?f=%5Cbeta%3D%5Carctan%28a%2B%5Csqrt%7Ba%5E2%2B1%7D%29)
<span>
In our case
![a=\tan(\dfrac{-35*2\pi}{360})=-\tan(\dfrac{7\pi}{36})](https://tex.z-dn.net/?f=a%3D%5Ctan%28%5Cdfrac%7B-35%2A2%5Cpi%7D%7B360%7D%29%3D-%5Ctan%28%5Cdfrac%7B7%5Cpi%7D%7B36%7D%29)
hence
![b=\arctan\left(a+\sqrt{a^2+1}\right)=27.5^\circ](https://tex.z-dn.net/?f=b%3D%5Carctan%5Cleft%28a%2B%5Csqrt%7Ba%5E2%2B1%7D%5Cright%29%3D27.5%5E%5Ccirc)
, which is the optimum angle to get a maximum range.</span></span>