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lisa tulip Grade: Upto college level
        

A particle of mass 2m is projected at an angle 45 degree with the horizontal with a velocity of 2root2 m/s. after 1 s explosion takes place and the particle is borken into two equal pieces. as a result one part comes to rest. find the maximum height attained by the other part.

7 years ago

Answers : (1)

Ramesh V
70 Points
										

The time of flight = 2u.sin x /g


                               =2*2.828*0.707 /10


                              =0.4 seconds


Making correction to 0.1 seconds, where explosion takes place


before explosion,


Vx =Vy = 2m/sec


before 0.1 sec,  Vx = 2 m/sec


                         Vy = 2 -0.1*10 = 1 m/sec


                         hz. distance covered = 2*0.1 = 0.2 m


                         vt.. distance covered = 2*0.1 - 5*0.01 = 0.15mts


before 0.1 sec,  explosion takes places


hz. and vt. moments shld be balanced which gives


in X dirn:    2m(2) = m(Vx' )


in Y dirn:     2m(1) = m(Vy' )


so, Vx' = 4 m/sec  and Vy' =2 m/sec


now, max. height attained is hmax = (Vy' )2 /2g


                                                              =4/20 = 0.2 m


so max. height attained from grnd level = 0.15 + 0.2 = 0.35 mts


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Regards,

Naga Ramesh

IIT Kgp - 2005 batch


 

7 years ago
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