A child stands on the edge of the cliff 10m above the ground and...

Given here, $u = 5m{s}^{- 1}, h = 10m$

Using ${v}_{y}^{2} = {u}_{y}^{2} + 2gh$ in vertical direction, we have the velocity of stone in vertical direction as ${v}_{y} = \sqrt{2gh} = \sqrt{200}m{s}^{- 1}$ (since, ${u}_{y} = 0$).

The final velocity along the horizontal direction is the same as the initial horizontal velocity as there is no acceleration along the x-axis.

The speed with which the stone hits the ground is

${v}_{net} = \sqrt{{v}_{x}^{2} + {v}_{y}^{2}} = \sqrt{25 + 200} = 15m{s}^{- 1}$