A particle is dropped from top of tower. During its motion it covers `(9)/(25)` part of height of tower in last 1 sec. Then find height of the tower.

Explanation

Let it taken n seconds to fall from tower then total distance `s = (1)/(2) g(n)^(2)`
distance travelled in `n^(th)` second
`s_(n^(th))= O + (g)/(2) (2n-1) rArr s_(n^(th)) = (9)/(25) s `
`rArr (g)/(2) (2n -1) = (9)/(25) (g)/(2)(n^(2)) rArr 9n^(2) = 50 n - 25 `
`rArr n = (50 + sqrt(50^(2) - 4 xx 9xx (+ 25)))/( 2xx 9) = 5 sec`.
`H = (1)/(2) g(5)^(2) = 125 `m