# Platformulae Solution

The following is a solution to Puzzle #9 – Platformulae. Take a look if you haven’t already.

It is given that the position of the platform goes like

$\bold{r}(t)=\sin\omega t\ \bold{i}$

This means that upon launch, your instantaneous velocity is

$\bold{v}=(v_0+\omega\cos\omega t)\bold{i}+(v_0)\bold{j}$

from the addition of your rest launch velocity and the transverse velocity of the platform at time $t$. We can now solve for the distance travelled from the point of launch. Your height above the platform $h$ as a function of time is

$h(t')=v_0 t'-\frac{1}{2}gt'^2$

from simple kinematics, where $g$ is the magnitude of acceleration due to gravity. We are using a separate, dummy time $t'$ so as not to confuse it with the time of launch $t$. The time at which you land $t_0$ satisfies

$h(t_0)=0$

for

$t_0\ne0$

Rearrangement yields

$\displaystyle t_0=\frac{2v_0}{g}$

Then the horizontal distance travelled is simply the product of time for which you travel, and your transverse velocity:

$\displaystyle \Delta x_1=(v_0+\omega\cos\omega t)\frac{2v_0}{g}$

We must also take into account the distance you have already been carried by the platform:

$\Delta x_2=\sin\omega t$

giving a total transverse displacement from the origin of

$\displaystyle \Delta x=\sin\omega t+(v_0+\omega\cos\omega t)\frac{2v_0}{g}$

We maximise this with respect to the launch time $t$:

$\displaystyle \frac{d\Delta x}{dt}=\omega\cos\omega t-\omega^2\sin\omega t\frac{2v_0}{g}$

which can be simplified to

$\displaystyle \tan\omega t=\frac{g}{2\omega v_0}$

At this point it becomes clear that I phrased the question poorly in the previous post. The question was when should you launch so as to travel the greatest distance. The implication was that ‘when’ was a spatial measure rather than a temporal one; that is: “how far from the origin are you when you jump?”

The transverse position at which you jump can be found by substituting the launch time $t$ into the expression for the position vector of the platform:

$x=\sin\omega t$

$x=\sqrt{1-\cos^2\omega t}$

$\displaystyle x=\sqrt{1-\frac{1}{\sec^2\omega t}}$

$\displaystyle x=\sqrt{1-\frac{1}{1+\tan^2\omega t}}$

$\displaystyle x= \frac{\tan\omega t}{\sqrt{1+\tan^2\omega t}}$

$\displaystyle x(\omega)=\frac{\frac{g}{2\omega v_0}}{\sqrt{1+\big(\frac{g}{2\omega v_0}\big)^2}}$

$\displaystyle x(\omega) = \frac{g}{\sqrt{(2\omega v_0)^2+g^2}}$

This is our result.

The result is a curve which is sharply peaked around the origin and decays rapidly for increasing $\omega$. Let’s interpret the results.

For $\omega=0, x=1$. Physically, $\omega\to0$ corresponds to a platform which moves at an infinitesimal speed. Clearly, there is no point in jumping before the platform has stopped moving, since the transverse speed boost you receive from it is almost imperceptible. Therefore it is most sensible to wait until it has stopped, and then jump.

For $\omega\to\infty, x\to0$. Physically, if the platform is travelling extremely quickly at its outset, you had best jump off immediately! The additional distance you would gain by jumping off later is negligible compared to that from the platform’s colossal initial speed.

The neat result is that there is a continuum of optimum launch positions as a function of varying $\omega$, which has a nice closed form and is easy to interpret.