Pessimist Singularitarian

#2
December 18th, 2016,
18:26
I would begin by orienting the vertical axis of motion such that point $p$ is at the origin. The first particle's position is then given by:

$ \displaystyle x_1=-\frac{a}{2}t^2+v_0t$

And the second particle's position is:

$ \displaystyle x_2=-\frac{a}{2}t^2+q$

Now, when the two particles meet, we have:

$ \displaystyle x_1=x_2\implies t=\frac{q}{v_0}$

At this time, their speeds are equal, hence:

$ \displaystyle -at+v_0=at\implies v_0=2at$

Hence:

$ \displaystyle t^2=\frac{q}{2a}$

And so we find:

$ \displaystyle r=-\frac{a}{2}\cdot\frac{q}{2a}+q=\frac{3}{4}q$

And this implies:

$ \displaystyle \overline{pr}=3\overline{rq}$