A hiker stands on the edge of a cliff 490   m above the ground and throws a stone horizontally with an…

From the equation of kinematics, $v^2-u^2=2gh$

There is no opposing force in the horizontal direction, $\Rightarrow v_x=15 \mathrm{m} {\mathrm{s}}^{-1}$

In vertical direction, ${v_y}^2=2gh$

$\Rightarrow {v_y}^2=2\times 9.8\times 490$

$\Rightarrow v_y=\sqrt{2\times 9.8\times 490}$

$\Rightarrow v_y=\sqrt{2\times 9.8\times 490}$

$\Rightarrow v_y=98 \mathrm{m} {\mathrm{s}}^{-1}$

Final velocity, $v=\sqrt{{v_x}^2+{v_y}^2}$

$\Rightarrow v=\sqrt{{15}^2+{98}^2}$

$\Rightarrow v=99 \mathrm{m} {\mathrm{s}}^{-1}$