A child stands on the edge of the cliff 10   m above the ground and throws a stone horizontally with an…

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

Using $v_y^2=u_y^2+2gh$ in vertical direction, we have the velocity of stone in vertical direction as $v_y=\sqrt{2gh}=\sqrt{200} \mathrm{m} {\mathrm{s}}^{-1}$   (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 

$v_{net}=\sqrt{v_x^2+v_y^2}=\sqrt{25+200}=15 \mathrm{m} {\mathrm{s}}^{-1}$