Motion under gravity : Revisited

For the first particle (which is dropped freely) initial velocity is zero and hence the displacement in time t is given as
$S = \frac{gt^2}{2}$
Putting S = H/n
$\frac{H}{n} = \frac{gt^2}{2}$
Solving for t
$t = \sqrt{\frac{2H}{ng}}$
For the second particle, Let the initial speed be u m/s
$\frac{(n-1)H}{n} = u\sqrt{\frac{2H}{ng}} - \frac{g}{2}(\frac{2H}{ng})$
Solving we get
$u = \sqrt{\frac{ngH}{2}}$

Now the velocity of first particle is
$v_{1} = g \sqrt{\frac{2H}{ng}}$
and the velocity of second particle is $v_{2} = \sqrt{\frac{ngH}{2}} - g \sqrt{\frac{2H}{ng}}$ We get $v_{1} : v_{2} = 2: (n-2)$