**This problem can be solved by using the n**^{th} second formula i.e. , S_{nth} = u + 1/2 a(2n-1) . Snth is the distance travelled in n^{th }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/s**^{2 } (as it is a freely falling body) we get two equations i.e.,

**S**_{nth} = 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/s**^{2} and t=3s in the equation S = ut + 1/2 gt^{2}, We get Height of the Tower ,S =45 m .

**If you are satisfied,then do give me thumbs up ,dude !!!**

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