A body falling from rest has a velocity 'V' after it falls through a distance 'h' . The distance it has to fall down further for its velocity to become double is …. Times 'h'.

A body falling from rest has a velocity 'V' after it falls through a d

1.	A body falling from rest has a velocity 'V' after it falls through a distance 'h' . The distance it has to fall down further for its velocity to become double is …. Times 'h'.
Answer» Solution :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`.