At time $t0 = 0$, a particle $P1$ of mass $m$ is projected vertically from the ground with speed $v0 > 0$. This particle reaches its maximum height $h1$ at time $t = t1$. At time $t2 > t1$, when this particle has the speed $v0/3$, another particle $P2$ of mass $3m$ is projected vertically upwards from the same point of the ground with speed $V0$. Assume that the gravitational acceleration $g$ is constant. (i) Determine $t1$ and $h1$. (ii) Determine $t2$ and the height $h2$ of $P1$ at $t$ = $t2$. (iii) Determine the time $t3$ and height $h3$ of the collision between particles $P1$ and $P2$
So the question (iii) I am struggling with.
I have calculated the following: $t1 = \frac{V0}{g}$
$h1 = \frac{V0^2}{2g}$
$t2 = \frac{4V0}{3g}$
$h2 = \frac{4V0^2}{9g}$ where $g$ is the constant gravity.
I have used the SUVAT formulae to try to work out the time of collision by assuming the displacement is the same for both particles, and putting $uv+0.5at^2$ of particle $1 = uv+0.5at^2$ of particle 2. However Im not sure i'm correct in doing this.
Any advice is appreciated (and sorry if this is long-winded, first time poster)