Question

a particle is dropped from the top of a tower.it covers 40m in last 2 seconds.find the height of the tower

Answer 1

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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