<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>
: mass of the ball
</span><span>
: angle by which the ground is inclined (here
)
</span><span>
: initial angle of the throw
</span><span>
: acceleration due to gravity,
</span><span>
: vertical unit vector, pointing upwards
</span><span>
: horizontal unit vector, pointing to the right
</span><span>
: x,y-position of the ball
</span><span>
: ball's acceleration
</span><span>
: ball's speed
</span><span>
: ball's position
</span><span>
: 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
.</span><span> By Newton's second law,
hence by integrating with respect to time,
which by integrating once more yields
</span><span>
Thus
. That last one gives
hence
</span><span>
The ball stops when it intersects with the slope defined by
hence if we call
the abscissa of the the intersection point,
hence
where
.
</span><span>
To find the maximum of
, we derive the function and solve
which is equivalent to
. Using the well known formula
we get
. The discriminant of this equation is
hence the unique positive solution is
thus
<span>
In our case
hence
, which is the optimum angle to get a maximum range.</span></span>