1. a particle is projected vertically upwards from a point P. at the same instant a second particle is let fall from rest vertically at q. q is directly above p. the 2 particles collide at a point r after t seconds . when the 2 particles collide they are travelling at equal speeds. prove that |pr|=3|rq|

i am trying to solve this with uvast equations for the first particle i have v=v s=r a=-g t=t for the second particle i have v=v s=p-r a=g t=t dont know where to go from here

2. 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:

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