A stone is thrown horizontally with velocity ‘g’ $m{s}^{-1}$ from the top of a tower of height ‘g’ meter. The speed with which it hits the ground is (in $m{s}^{-1}$)

$$\begin{gathered} final velocity =v=v_{x}i+v_{y}j \\ magnitude of velocity=\left|v\right|=\sqrt{{v_x}^2+{v_y}^2} \\ {v}_x=u=g=horizontal componet of velocity is cons\tan t. \\ from {v_y}^2={u_y}^2+2gh; here u_y=0 \\ {V}_y=\sqrt{2gh} ; substitute given value \\ \Rightarrow V_y=\sqrt{2g\times g} \\ substitute in \left|v\right|=\sqrt{{v_x}^2+{v_y}^2} substitute the obtained values, \\ \Rightarrow V=\sqrt{g^2+2g^2} \\ \Rightarrow V=\sqrt{3}g \end{gathered}$$