Question

a stone dropped from the top of a tower is found to travel 5/9 of the height of the tower during the last second of its fall. the time of fall is?

Answer 1

let the height of tower be x
u = 0m/s
so let time to cover 5/9x of building be t sec
so time taken to cover 4x/9 of the building = t - 1 sec
$S_{t} = u + 1/2g(2t - 1)$
 5x/9= 5(2t - 1) ...........(i)  g = 10
so 
again s = ut +1/2gt²
4x/9 = 0 + 5(t - 1)² ..................(ii)
 dividing i and ii
 5x/9/4x/9  =5(2t - 1) /5(t - 1)²
⇒5t² +5 - 10t = 8t - 4
⇒5t² -18t +9 =0
solving t we get 
t = 3 sec and 0.6 sec

Answer 2

if total time of fall is t
h=1/2 gt²
fall in t-1 time period = h- 5/9h= 4/9h
Thus, 4/9h = 1/2 g (t-1)²
 dividing first equation by the second                
t²/(t-1)² = 9/4
taking square root
t/(t-1)= 3/2
2t=3t-3
t=3 (answer)