Question

particle is dropped from a tower. it covers 40m in last 2s. Find the hight of the tower

Answer 1

s = u + g/2 (2n-1)

s = height in any second 
u = starting velocity
g = gravitational acceleration
n = height cover in that second

Let the total time taken  by the body to reach the bottom be n seconds.
So, the last two seconds will be n and (n-1).
So, by putting the values   
u=0 m/s and g=10m/s (as it is a freely falling body)
we get two equations i.e.,

S = 0 + 1/2 g(2n-1)  = 5(2n-1)  .  ..........................1st equation
S = 0 + 1/2 g{2(n-1)-1}  =5(2n-3)  ..........................2nd equation
from 1 and 2 , given, equation 1+equation 2 = 40 m. 
So add both the equations
5(2n-1) +5(2n-3) =40 
By solving this we get n=3
( Hence total time ,t = 3 seconds)
 By putting the values of u=0m/s ,g=10 m/s and t=3s  in the equation 
 S = ut + 1/2 gt^2 ,
We get Height of the Tower ,S =45 m . 

i hope it will help you 
regards