A stone is thrown vertically upward with a speed of 10.0 `ms^(-1)` from the edge of a cliff 65 m high.How much later will it reach the bottom of the cliff ? What will be its speed just before hitting the bottom ?

Explanation

u=10 m/s , s=65 m
we know that `s=ut+(1)/(2)at^(2)`
`-65 10t-(1)/(2)9.8 t^(2)-65=10t-4.9t^(2)-10t-65=0`
`t=(10pmsqrt(100+4xx4.9xx65))/(2xx4.9)=(10pmsqrt(100+1274))/(9.8)`
`=(10pmsqrt(1374))/(9.8)=(10pm37.06)/(9.8)`
`=(47.06)/(0.8)=4.8` sec, speed of the stone
`v=sqrt(u^(2)+2gh)=sqrt(2xx9.8xx6.5+100)`
`=sqrt(19.6xx65+100)=sqrt(1374)=37m/s`