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A body falling from the rest has a veloc...

A body falling from the rest has a velocity v after it falls through a heigh h. The distance it has to fall down further for its velocity to become double, will be

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Let a body starting from rest travels a distance of .h. m from A to B during which it acquires a velocity V as shown in the figure . Its velocity becomes 2V at point C.

`v^(2)-u^(2)=2as`
Apply the above formula for both the cases we get,
`V^(2)-0^(2)=2gh` `(2V)^(2)-V^(2)=2gh`
`rArr V^(2)=2gh` .....(1)
`rArr 3V^(2)=2gh^(1)` .....(2)
Eliminating the unconcerned terms by dividing equation (2)by equation (1) we get,
`(2)/(1)=(3V^(2))/V^(2)=(2gh^(1))/(2gh)rArrh^(1)=3h`.
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