A hiker stands on the edge of a cliff `490 m` above the ground and throwns a stone horiozontally with an initial speed of `15ms^(-1)` neglecting air resistance.The speed with which it hits the ground in `ms^(-1)` is `(g=9.8ms^(2))`

Explanation

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 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_(@)+v_(@x)t`
`y(t)=y_(@)+v_(@y)t+(1//2)a_(y)t^(2)`
Here , `x_(@)=y_(@)=0,v_(@y)=0,a_(y)=-r=-9.8 " m s "^(-2),v_(@x)=15 " m s"^(-1)`.
The stone hits the ground when y(t) =- 490 m.
`-490m=-(1//2)(9.8)t^(2)`.
This gives `t=10s`.
The velocity components are `v_(x)=v_(@x)andv_(y)=v_(@y)- gt`
so that when the stone hits the ground :
`v_(@y)=15 " m s"^(-1)`
`v_(@y)=0-9.8xx10=-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)`