# 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.

Ramesh V
70 Points
12 years ago

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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