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Q.

A particle  of mass 2m is projected at an angle of 450 with horizontal with a velocity of 202 m/s. An explosion takes place after 1 s and the particle is broken into two equal pieces. As a result of explosion one particle comes to rest. The maximum height from the ground attained by the other part is (g=10 m/s2

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a

20 m

b

30 m

c

35 m

d

40 m

answer is C.

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Detailed Solution

Height at which explosion occurs H=(usinθ)t12gt2=15m
Just before explosion 
Vx=ucos θ=20ms1Vy=(usin θgt)=10ms1
Applying LCLM,
V2=40i^+20j^h=V2Y22g=20m
Maximum height reached by the fragment= H+h=35m 

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