Max throw
At what angle should you throw something to maximise the distance it travels?
Problem
A particle is projected with speed $10 \mathrm{m s}^{-1}$ from a flat horizontal surface. Find, with proof, the angle from which it should be projected to maximise the distance travelled before it hits the surface. Does this angle depend on the speed of projection?
The particle is now projected with speed $10 \mathrm{m s}^{-1}$ from a height of $2$ metres.
From what angle (to 3sf) should it now be projected to maximise the distance travelled before it hits the surface? Does this angle depend on the speed of projection?
Did you know ... ?
The modelling assumptions of constant gravitational field and no friction opposing motion are good ones, leading to simple equations which always have parabolas for solution. Once these modelling assumptions are, rightly, challenged, the resulting equations become 'non-linear' and very difficult to solve. Mathematicians often take the parabola as a starting point to solving the more complicated equations and vary the solution a little to try to fit it back into the new equations. You can see an aspect of this process in the solution to this problem.
Student Solutions
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
$$
(x, y) = \left(V \cos(\alpha) t, -0.5 g t^2+V\sin(\alpha) t +H\right)\;.
$$
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 $100\mathrm{ms}^{-1}$. 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
$$
\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right) + \sin(\alpha)\cos(2\alpha)\left(1 +\frac{2gH}{V^2\sin^2(\alpha)}\right)^{\frac{1}{2}}=0\;.
$$
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 $\cos(\alpha)\ldots$
Teachers' Resources
Why do this problem?
This problem starts off as a reasonably straightforward application of kinematics in the context of projectile motion; the interest lies in comparing the well-understood situation where the particle is projected from ground level with the situation where the point of projection is just two metres above the ground. Both scenarios involve trig identities, but the second scenario can be approached with advanced differentiation or numerically.
Possible approach
Students could be asked for a "gut reaction" to the initial projection from ground level problem and also an intuitive explanation to back this up. Then they can be encouraged to prove their result mathematically.
Students can then think about how to modify their approach to the first problem when projection is from $2$ m above ground. There are multiple ways to do this, such as adding $2$ on to their expression for the $y$-coordinate of the particle or taking the ground level to be $y=-2$. At this point, they could consider how they would go about finding the optimal angle and whether this is feasible. If time allows, a spreadsheet could be used to appromximate the optimal angle for differnet initial speeds.
Possible extension
$$
\frac{g}{V^2}\frac{dx}{d\alpha} = \cos(2\alpha)-\sin(\alpha)\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}+\cos^2\alpha\sin\alpha \frac{1}{\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}}\;.
$$
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 $\cos(\alpha)\ldots$
Possible support
A simulation like https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_all.html can be used to help students see the differences in ranges according to angle and height of projection for different initial speeds.