A body is projected horizontally from a height of 5 m. It reaches the ground at a horizontal distance of 10 m. The speed of the particle when it reaches the ground is $\left(\mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2\right)$

Time taken to reach the ground is $\mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}=\sqrt{\frac{2\times 5}{10}}=1 \mathrm{s}$

Speed of projection is $\mathrm{u}=\frac{\mathrm{R}}{\mathrm{t}}=\frac{10}{1}=10 \mathrm{m}/\mathrm{s}$

The horizontal component of velocity on reaching the ground is 

${\mathrm{\nu }}_{\mathrm{x}}=\mathrm{u}=10 \mathrm{m}/\mathrm{s}$

The vertical component of velocity on reaching the ground is 

${\mathrm{\nu }}_{\mathrm{y}}=\sqrt{2\mathrm{gh}}$

$=\sqrt{2\times 10\times 5}=10 \mathrm{m}/\mathrm{s}$

Speed on reaching the ground $=\sqrt{{\mathrm{\nu }}_{\mathrm{x}}^2+{\mathrm{\nu }}_{\mathrm{y}}^2}=\sqrt{{\left(10\right)}^2+{\left(10\right)}^2}=10\sqrt{2} \mathrm{m}/\mathrm{s}$