A slide with a frictionless curved surface, which becomes...

Using energy conservation,

$mgh = \dfrac{1}{2}m{\left({u}_{0}\right)}^{2}$

$⇒ {u}_{0} = \sqrt{2gh}$

${u}_{0}$ is the horizontal component of the velocity throughout the flight. For vertical component of the velocity(just before collision with the ground), we can use equation of motion

${\left({v}_{z}\right)}^{2} = {0}^{2} + 2\left(- g\right)\left(- 3h\right) ⇒ {v}_{z} = \sqrt{6gh}$

Angle $\theta$ can be written as,

$tan\theta = \dfrac{{v}_{z}}{u} = \sqrt{3}$

$⇒ \theta = {60}^{o}$

Horizontal distance covered just before the collision will be,

$d = {u}_{0}T = {u}_{0}\sqrt{2\left(\dfrac{3h}{g}\right)} = \sqrt{2gh}\sqrt{\left(2\right)\left(\dfrac{3h}{g}\right)} = 2\sqrt{3}h$

After collision, only velocity along $z$-direction changes. Therefore,

${v}_{1} = e{v}_{z} = \sqrt{2gh}$ and hence

$\overset{\to}{v} = {u}_{0}\overset{∧}{i} + {v}_{1}\overset{∧}{k}$

$= \sqrt{2gh}\left[\overset{∧}{i} + \overset{∧}{k}\right] = \sqrt{2gh}\left[\overset{∧}{x} + \overset{∧}{z}\right]$

Now, maximum height achieved after collision will be,

${h}_{1} = \dfrac{{{v}_{1}}^{2}}{2g} = h$.

Finally, ${u}_{0} = \sqrt{2gh}, \theta = {60}^{o}, \dfrac{d}{{h}_{1}} = 2\sqrt{3}$.