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A stone is thrown vertically upward with...

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 ?

Text Solution

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`u=10m//s, s=65m`
we know that `s=ut+1/2at^(2)`
`-65=10t-1/29.8t^(2)-65=10t-4.9t^(2)-10t-65=0`
`t=(10+-sqrt(100+4xx4.9xx65))/(2xx4.9)=(10+-sqrt(100+1274))/(9.8)`
`=10+-sqrt(1374)/(9.8)=(10+-37.06)/(9.8)`
`t=(47.06)/(9.8)=4.8 sec, ` speed of the stone
`v=sqrt(u^(2)+2gh)=sqrt(2xx9.8xx65+100)`
`=sqrt(19.6xx65+100)=sqrt(1374)=37m//s`
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