A particle is released from rest from a tower of height 3h. The ratio of the intervals of time to cover three equal height h is.

Options

A. `t_(1):t_(2):t_(3)=3:2:1`
B. `t_(1):t_(2):t_(3)=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

`t_(1)` - Cover `h_(1)` distance
`t_(2)` - Cover `h_(2)` distance
`t_(3)`- Cover `h_(3)` distance
Now for S = `2`h g=g u=O …(i)

`t_(1)=sqrt((2h)/(g))`
`:.2h=(1)/(2)g(t_(2))^(2)`
`t_(2)=2sqrt((h)/(g))`
`:.t_(2)=t_(2)-t_(1)`
`=2sqrt((h)/(g))-sqrt((2h)/(g))=sqrt((2h)/(g))(sqrt(2)-1)`
`t_(2)=sqrt((2h)/(g))(sqrt(2)-1)` ...(ii)
For S=`3` h, u= `0` g=g
`t_(3)=sqrt((6h)/(g)),t_(3)=t_(3)-t_(2)-t_(1)`
`=sqrt((2h)/(g))(sqrt(3)-sqrt(2)+1-1)`
`t_(3)=sqrt((2h)/(g))(sqrt(3)-sqrt(2))`
`t_(1):t_(2):t_(3)=1:(sqrt(2)-1):(sqrt(3)-sqrt(2))`