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

D

Explanation

`Deltak_(1) = (1)/(2) mg^(2) ((t^(2))/(9)) (because Delta k = (1)/(2) mv^(2) ,v = g t)`
`Delta k_(2) = (1)/(2) mg^(2) (t^(2) -(4t^(2))/(9))`.