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 time taken by the stone to reach the ground in seconds is `(g=9.8ms^(2))`

Explanation

The stone is thrown from the edge for the cliff with a speed `u=15 m//s`, horizontally. It hits the ground after time t .
Considering the vertically downward motion, we have
`h=u_(y)t+(1)/(2)"gt"^(2)=0xxt+(1)/(2)xxgxxt^(2)`
`rArr t^(2)=(2h)/(g) or t=sqrt((2h)/(g))=sqrt((2xx490)/(9.8))=sqrt100`
`t=10s`
`[ :.u_(y)=0` initially]
Horizontal component of final velocity , `u_(x)=u=15m//s`
Vertical component of final velocity
`v_(y)=u_(y)t+"gt"=0+9.8xx10=98m//s`