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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.
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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