A particle is droped under gravety from rest a height h and it travels a distance `(9h)/(25)`in the last second, the height h is (take `g=9.8ms^(-2)`)

Explanation

`(h)nth=u+(2n+1)/(2)g` (or) `(9h)/(25)=0+(2n-1)/(2) g` …….(i)
`h=0+(1)/(2)gxx(n)^(2)` ……..(ii)
(i)/(ii), we `(9)/(25)=(2n-1)/(n^(2))`
9 `n^(2)`=50 n-25
9 `n^(2)`-50 n+25=0
After solving n =5s,`(5)/(9)s`
H=122.5 m, 1.5 m.