A,B,C and D are four points on a vertical line such that AB = BC CD. A body is allowed to fall freely from A. Prove that the respective times required by the body to cross the distances AB, BC, CD should be in the ratio `1 : (sqrt(2)-1) : (sqrt(3)-sqrt(2)).`

Explanation

Let AB = BC = CD = x and time taken by the body to cover these distances be`t_(1), t_(2)` and `t_(3)` respectively.
Now, `x = (1)/(2)"gt"_(1)^(2) " ""or", t_(1) = sqrt((2x)/(g)) " "cdots(1)`
`2x = (1)/(2) g(t_(1)-t_(2))^(2) " ""or", t_(1)+t_(2)= sqrt((4x)/(g)) " "cdots(2)`
`3x = (1)/(2) g(t_(1)+t_(2)+t_(3))^(2)" ""or" t_(1)+t_(2)+t_(3)= sqrt((6x)/(g)) " " cdots(3)`
From (1) and (2) we get,
`t_(2) = sqrt((4x)/(g))-sqrt((2x)/g)=sqrt((2x)/(g))(sqrt(2)-1)`
From (2) and (3) we get ,
`t_(3)=sqrt((6x)/(g))-sqrt((4x)/g)=sqrt((2x)/(g))(sqrt(3)-sqrt(2))`
`:. " " t_(1):t_(2):t_(3)=1:(sqrt2-1):(sqrt3-sqrt(2))" "` (Proved).