Strange Kinematic Optimization

Anticipating dealing with lots of occurrences of $\frac{2}{3}\theta$ later on, put $\theta=3\psi$ .

The trajectory of the drone is given by

$\begin{aligned} x&=vt\cos{3\psi} \\ y&=vt\sin{3\psi}-\frac{gt^2}{2} \end{aligned}$

There are various standard ways to find the highest point, which we'll say has coordinates $(p,q)$ . We can consider the horizontal and vertical components of the drone's velocity, which are given by

$\begin{aligned} \dot x&=v\cos{3\psi} \\ \dot y&=v\sin{3\psi}-gt \end{aligned}$

At the highest point, $\dot y=0$ ; which happens when $t=\frac{v\sin{3\psi}}{g}$ , so that

$\begin{aligned} p&=\frac{v^2}{2g} \sin{6\psi} \\ q&=\frac{v^2}{g}\sin^2{3\psi}-\frac{v^2}{2g}\sin^2{3\psi}=\frac{v^2}{2g}\sin^2{3\psi} \end{aligned}$

Let's tidy this up a bit more. We can choose $v$ , so let's put $v^2=2g$ ; then we get

$\begin{aligned} p&=\sin{6\psi} \\ q&=\sin^2{3\psi} \end{aligned}$

Now, we need to consider the projectile shot by the drone. Let's use $\tau$ to measure time from this point (so that the projectile is fired at $\tau=0)$ . We are told the projectile is fired at a speed $v$ and angle $\frac{2}{3}\theta=2\psi$ relative to the drone.

At its highest point, the vertical component of the drone's velocity is zero; the horizontal component is $v\cos{3\psi}$ . Hence the initial velocity of the projectile is $(v\cos{3\psi}+v\cos{2\psi},v\sin{2\psi})$

The trajectory of the projectile is

$\begin{aligned} x&=p+v(\cos{3\psi}+\cos{2\psi})\tau \\ y&=q+v\tau\sin{2\psi}-\frac{g\tau^2}{2} \end{aligned}$

To find the overall range, $R$ of the projectile, we need to find $x$ when $y=0$ . Say this happens when $\tau=T$ .

$\begin{aligned} 0&=q+vT\sin{2\psi}-\frac{gT^2}{2} \\ T&=\frac{v\sin{2\psi} \pm \sqrt{v^2\sin^2{2\psi}+2gq}}{g} \end{aligned}$

We want the positive root. Substituting in,

$\begin{aligned} R&=p+v(\cos{3\psi}+\cos{2\psi})\frac{v\sin{2\psi} + \sqrt{v^2\sin^2{2\psi}+2gq}}{g} \\ &=p+2(\cos{3\psi}+\cos{2\psi})(\sin{2\psi} + \sqrt{\sin^2{2\psi}+q}) \end{aligned}$

where again we've made use of our choice $v^2=2g$ . Plugging in $p$ and $q$ ,

$\begin{aligned} R&=\sin{6\psi}+2(\cos{3\psi}+\cos{2\psi})(\sin{2\psi} + \sqrt{\sin^2{2\psi}+\sin^2{3\psi}}) \end{aligned}$

This can be maximised either numerically (directly) or by setting the derivative equal to zero and solving that numerically. Either way gives $\psi\approx17.1638^\circ$ , $\theta\approx51.4914^\circ$ and the answer $\lfloor 100\theta' \rfloor=\boxed{5149}$