If in the previous problem, the particle travels `9//25` of the distance in its last second, find the time of fall and the height of the tower.

Explanation

`h=1/2g t_(0)^(2) …(i)`
`(9h)/25=1/2g(2t_(0)-1) …(ii)`
`(9)/(25).(1)/(2) g t_(0)^(2)=1/2g(2 t_(0)-1)`
`9t_(0)^(2)-50t_(0)+25=0`
`t_(0)=(-(-50)+-sqrt((-50)^(2)-4xx9xx25))/(2xx9)`
`=(50+-40)/18`
`implies t_(0)=5 s, t_(0)=5/9 s`
`t_(0)=5/9 s` is not possible
`h=1/2g t_(0)^(2)=1/2xx10(5)^(2)=125 m`