Answer we choose the origin of the $x$ – and $y$ axis at the edge of the cliff and $t=0$ s at the instant the stone is thrown. Choose the positive direction of $x$ -axis to be along the initial velocity and the positive direction of $y$ -axis to be the vertically upward direction. The $x-,$ and $y$ components of the motion can be treated independently. The equations of motion are:

$x(t) =x_{o}+v_{\alpha x} t$

$y(t) =y_{o}+v_{o y} t+(1 / 2) a_{y} t^{2}$

Here, $x_{0} =y_{0}=0, v_{o y}=0, a_{y}=-g=-9.8 m s ^{-2}$

$v_{ox }=15 m s ^{-1}$

The stone hits the ground when $y(t)=-490 m$

$-490 m =-(1 / 2)(9.8) t^{2}$

This gives $t=10 s$ The velocity components are $v_{x}=v_{a x}$ and $v_{y}=v_{o y}-g t$

so that when the stone hits the ground:

$v_{o x}=15 m s ^{-1}$

$v_{o y}=0-9.8 \times 10=-98 m s ^{-1}$

Therefore, the speed of the stone is

$\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{15^{2}+98^{2}}=99 m s ^{-1}$