Gravity Projectile again

As the principal of conservation of energy says that the loss in potential energy is equal to gain in kinetic energy for a system so the speed of the object remains the same only the direction changes . So the velocity of the object at the height H is ( let it be ) = v and as per given information the velocity below point of projection is 2 v , and let the initial projection velocity = u .

Now we have velocity above point of projection at H height as $v^{2} = u^{2} - 2gH$ and we have velocity below point of projection at H depth as $(2v)^{2} = u^{2} +2gH$

Solving these two equations we get $3v^{2} = 4gH$

and so we have $\frac{5}{2}v^{2} = u^{2}$

and at highest point of the trajectory for the particle the velocity is zero and so $u^{2} = 2g(maximum height)$ $maximum height = \frac{u^{2}}{2g}$

which after calculation gives

$maximum height = \frac{5H}{3}$

so $n = 5$