A body falling from rest has a velocity v after it falls through a distanc.e h. The distance it has to fall down further, for its velocity to become double, is ..... times h

Explanation

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)=2 as`
Apply the above formula for both the cases we get,
`V^(2)-0^(2)2gh" "(2V)^(2)V^(2)=2gh^(1)`
`impliesV^(2)=2gh` ......(1)
`implies3V^(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)impliesh^(1)=3h`.