A particle is dropped from the top of a tower.
(a) Find the ratio of displacement in successive time interval `t_(0)`.
(b) Find the ratio of time in falling successive distances `h`.

Explanation

`O` to `A`: `h_(1)=1/2g t_(0)^(2) …(i)`
`O` to `B`: `h_(1)+h_(2)=1/2 g(t_(0)+t_(0))^(2)=4.1/2g t_(0)^(2)`
`h_(2)=(4-1)1/2g t_(0)^(2)`
`=3.(1)/(2)g t_(0)^(2) …(ii)`
`O` to `C`: `h_(1)+h_(2)+h_(3)=1/2g(t_(0)+t_(0)+t_(0))^(2)`
`=9.(1)/(2)g t_(0)^(2)`
`h_(3)=(9-4)1/2g t_(0)^(2)`
`=5.(1)/(2) g t_(0)^(2) ...(iii)`
`h_(1):h_(2):h_(3): ...=1/2g t_(0)^(2):3.(1)/(2)g t_(0):5.(1)/(2)g t_(0)^(2): ...`
`=1:3:5: ...`

Recall: If a particle is moving in a straight line under constant acceleration//deceleration, the ratio of displacement in successive time-intervals are in the ratio of add numbers.
(b) `O` to `A`: `h=1/2g t_(1)^(2)`
`t_(1)=sqrt((2h)/g) ...(i)`
`O` to `B`: `h+h=1/2g(t_(1)+t_(2))^(2)`
`t_(1)+t_(2)=sqrt((4h)/g)`
`t_(2)=sqrt((4h)/g)-sqrt((2h)/g) ...(ii)`
`O` to `C`: `h+h+h=1/2g(t_(1)+t_(2)+t_(3))^(2)`
`t_(1)+t_(2)+t_(3)=sqrt((6h)/g)`
`t_(3)=sqrt((6h)/g)-sqrt((4h)/g) ...(iii)`
`t_(1):t_(2):t_(3):....=sqrt((2h)/g):(sqrt((4h)/g)-sqrt((2h)/g)):(sqrt((6h)/g)-sqrt((4h)/g))...`
`=1:(sqrt(2)-1):(sqrt(3)-sqrt(2)) ...`