A body freely falls from a certain height onto the ground in a time 't'. During the first one third of the interval it gains a kinetic energy `Delta K_(1)` and during the last one third of the interval, it gains a kinetic energy `Delta K_(2)`. The ratio `Delta K_(1) : Delta K_(2)` is

Options

A. `1 : 1`
B. `1 : 3`
C. `1 : 4`
D. `1 : 5`

Answer

`(4)`

Explanation

`Delta k_(1) = 1/2mg^(2)((t^(2))/9)(therefore Delta k = 1/2 mv^(2) ( v = gt)`
`Delta k_(2) = 1/2 mg^(2) (t^(2) - (4t^(2))/9)` or `DeltaK = FS Rightarrow Delta K infty S`