An object is allowed to fall freely from a cliff. When it travels a distance 'h', its velocity is v. Hence, in travelling further distance of ........ its velocity will become 2v.

Options

A. 4h
B. 3h
C. 2h
D. h

Answer

C

Explanation

Equation for freely falling body,
`v ^(2) -v _(0)^(2) =2gd`
Here initial velocity `v _(0) = 0 and d =h`
`therefore v ^(2) =2gh`
` therefore h = (v ^(2))/(2g) " "…(1)`
The velocity of an object at height h is `v _(0) = v and ` the velocity of an object at height h. will be 2v.
`therefore (2v)^(2) -v ^(2) =2gh.`
`therefore 4v ^(2) -v ^(2) = 2gh.`
`therefore 3v ^(2) =2gh.`
`therefore h . = (3v ^(2))/( 2g) " "...(2)`
`therefore (h.)/(h) =3`
[Taking the ratio of equ. (2) and equ. (1)]
`therefore h. =3h`