This problem can be solved by using the nth second formula i.e. , Snth = u + 1/2 a(2n-1) . Snth is the distance travelled in nth second , u=initial velocity and a = g
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/s2 (as it is a freely falling body) we get two equations i.e.,
Snth = 0 + 1/2 g(2n-1) = 5(2n-1) . ..........................1st equation
S(n-1)th = 0 + 1/2 g(2(n-1)-1) =5(2n-3) ..........................2nd equation
from 1 and 2 , given, 1+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/s2 and t=3s in the equation S = ut + 1/2 gt2, We get Height of the Tower ,S =45 m .
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