Question

A particle dropped from the top of a tower. during its motion it covers 9/25 part of height of tpwer in the last 1 second Then find rhe hight of tower......???

Answer 1

Height of tower = H
Total time taken to reach the ground = T

16H/25 = 0.5gT^2--------(1)

And, discounting air resistance as nothing's been given for that, U = sqrt(2g(16/25)H) assuming the drop means no initial speed at the top. Note 16/25 H is the height the body dropped up to the last second. 

9H/25 = 0.5g (T - 1)^2--------(2)

Divide above two equations 
16 / 9 = (T / (T -1))^2
4 / 3 = T / (T - 1)
3T = 4T - 4
T = 4 s

Substitute T = 4s in equation (1)
16H / 25 = 0.5g * 4^2
H = (25/16) * 0.5 * 9.8 * 16
H = 122.5 m

Height of the tower is 122.5 m

i hope it will help you 
regards