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

 

$$(x,y)=(V\cos (\alpha )t,-0.5g{t}^2+V\sin (\alpha )t+H).$$

 

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

$$(x,y)=(\frac{2V^2}{g}\cos (\alpha )\sin (\alpha ),0)=(\frac{2V^2}{g}\sin (2\alpha ),0),$$ 

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

$$x=\frac{V^2}{g}\cos (\alpha )(\sin (\alpha )+\sqrt{{\sin }^2(\alpha )+\frac{2gH}{V^2}}).$$

 

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 $100\mathrm{ms}^{-1}$. 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

$$\sin \alpha (\cos (2\alpha )-\frac{2gH}{V})+\sin (\alpha )\cos (2\alpha ){(1+\frac{2gH}{V^2{\sin }^2(\alpha )})}^{\frac{1}{2}}=0.$$

If $X(\alpha) \equiv \frac{2gH}{V^2\sin^2(\alpha)}$ is small then we can expand to give

 $$\sin \alpha (\cos (2\alpha )-\frac{2gH}{V})+\sin (\alpha )\cos (2\alpha )(1+\frac{1}{2}\frac{2gH}{V^2{\sin }^2(\alpha )}+\mathcal{O}(X^2))=0.$$

In principle that can now be turned into a polynomial expression in $\cos(\alpha)\ldots$