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a body falls freely from rest covers more distances in each next second .these distances are in ratio?

Anandhi Selvi , 7 Years ago
Grade 12th pass
anser 2 Answers
Khimraj

Last Activity: 7 Years ago

Distance traveled in each nth sec is given bySn = (1/2)g(2n-1)So the ratio of distance in 1st sec 2nd sec 3 sec and so on is given by1:3:5:7:9...

Nikhil Sukhani

Last Activity: 7 Years ago

Let u = initial speed 
      t = time taken
      a = acc. 
then ,
 distance covered in n seconds (S1) = u(n) +1/2a(n)2
                                                                     = nu + 1/2an2
 distance covered in n-1 seconds (S2) = u(n-1) +1/2a(n-1)2
                                                                           = nu - u + 1/2an2 +1/2a - an 
  distance vovered in nth second = Sn – Sn-1
                                                              = nu +1/2an2 - ( nu - u +1/2an2 +1/2a – an )
                                               = u – 1/2a +an
                                               = u -a(1/2-n)
                                               = u + a/2(2n-1)
Since body falls freely from rest
Therefore ,  a=g
                  u=0
So , the formula becomes g/2(2n-1) 
So the ratio of distance travelled in 1st sec , 2nd sec 3rd sec.... is 1:3:5:...
CHEERS ;)
       
                                            
                                                

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