A particle is released from rest from a tower of height 3h. The ratio of time intervals for fall of equal height h i.e. `t_(1):t_(2):t_(3)` is :

Options

A. `t_1 : t_2 : t_3 = 3 : 2 : 1`
B. `t_1 : t_2 : t_3 = 1 : (sqrt 2 - 1) : (sqrt 3 - 2)`
C. `t_1 : t_2 : t_3 = sqrt 3 : sqrt 2 : 1`
D. `t_1 : t_2 : t_3 = 1 : (sqrt 2 - 1) : (sqrt 3 - sqrt 2)`

Answer

D

Explanation

Let `t_1, t_2, t_3` be the timings for three successive equal heights h covered during the free fall of the particle.
Then
`h = 1/2 gt1^(2)_1` or `t_1 = sqrt((2h)/g)` (since u = 0) .. (i)
`2h = 1/2 g(t_1 + t_2)^(2)` or `t_1 + t_2 = sqrt ((4h)/g)` ...(ii)
`3h = 1/2 g(t_1 + t_2 + t_3)^(2)` or `t_1 + t_2 + t_3 = sqrt((6h)/g)` ...(iii)
Subtracting (i) from (ii), we get
`t_2 = sqrt ((4h)/g) - sqrt ((2h)/g) = sqrt ((2h)/g)(sqrt 2 - 1)` ....(iv)
Subtracting (ii) from (iii), we get
`t_3 = sqrt ((6h)/g) - (4h)/g = sqrt ((2h)/g)(sqrt 3 - sqrt 2)`