Weekly Challenge 26: Max Throw
Suppose that the particle is projected from a height $H$ above
the ground at speed $V$ at an angle $\alpha$ to the $x$-axis, with
$x$ measuring the horizontal distance travelled and $y$ measuring
the vertical distance travelled.
Then the coordinates of the points along this trajectory are
$$
(x, y) = (V \cos(\alpha) t, -0.5 g t^2+V\sin(\alpha) t +H)
$$
The particle intersects the $x$-axis when the $y$-coordinate is
zero. This is when
$$
t = \frac{V\sin(\alpha) \pm \sqrt{V^2\sin^2(\alpha) +2gH}}{g}
$$
The particular case
$H=0$
The square root simplifies giving the point of intersection
as
$$
(x, y) = \left(\frac{2V^2}{g}\cos(\alpha)\sin(\alpha), 0\right) =
\left(\frac{2V^2}{g}\sin(2\alpha), 0\right)\,,
$$
where the second equality makes use of a trig identity. The
$x$-value $V\sin(2\alpha)$ is maximised when $2\alpha=90^\circ$.
Thus the optimal angle of projection is $45^\circ$, for any initial
velocity.
When $H\neq 0$
In this case, the particle intersects the $x$-axis at
$$
x =\frac{V^2}{g} \cos(\alpha) \left(\sin(\alpha)
+ \sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}\right)
$$
This looks complicated to differentiate so I tried a numerical
solution using a spreadsheet.This produced an optimal angle of
$40.4^\circ$ (3sf).
It seems clear that this angle will be dependent on the initial
speed, but to check I calculate the optimum angle numerically for a
large initial speed of $100$m$s^{-1}$. This produces an optimium of
$44.9^\circ$ (3sf) which is close to $45^\circ$, as we might
intuitively expect.
Extension:
You might like to investigate this further. Here are some starting
points.
Note that the expression for the derivative is:
$$
\frac{g}{V^2}\frac{dx}{d\alpha} =
\cos(2\alpha)-\sin(\alpha)\sqrt{\sin^2(\alpha)
+\frac{2gH}{V^2}}+\cos^2\alpha\sin\alpha \frac{1}{\sqrt{\sin^2(\alpha)
+\frac{2gH}{V^2}}}
$$
For an optimum we can set the left hand side to zero. This gives
us
$$
\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right)
+ \sin(\alpha)\cos(2\alpha)\left(1
+\frac{2gH}{V^2\sin^2(\alpha)}\right)^{\frac{1}{2}}=0
$$
If $X(\alpha) \equiv \frac{2gH}{V^2\sin^2(\alpha)}$ is
small then we can expand to give
$$
\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right)+
\sin(\alpha)\cos(2\alpha)\left(1+\frac{1}{2}\frac{2gH}{V^2\sin^2(\alpha)}+\mathcal{O}(X^2)\right)=0
$$
In principle that can now be turned into a polynomial expression in
$\cos(\alpha)$.....