Question

# A particle of mass $$2m$$ is projected at an angle of $${45}^{o}$$ with horizontal with a velocity of $$20\sqrt { 2 } m/s$$. After $$1s$$ explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take $$g=10m/{ s }^{ 2 }$$:

A
25m
B
50m
C
15m
D
35m

Solution

## The correct option is D $$35m$$Height moved by particle in $$1sec$$, $$h=ut-\cfrac { 1 }{ 2 } gt^{ 2 }$$$$=20\times 1-\cfrac { 1 }{ 2 } \times 10\times 1$$$$=15m$$After $$1sec$$ vertical speed is $$v$$, $$v-u=-gt$$$$v-20=-10\times 1$$$$v=10m/s$$ angle $$\theta$$ at height $$h$$ with horizontal is,$$\tan { \theta } =\cfrac { 10 }{ 20 } =\cfrac { 1 }{ 2 } ,\sin { \theta } =\cfrac { 1 }{ \sqrt { 5 } }$$before explosion velocity$$=\sqrt { { 10 }^{ 2 }+{ 20 }^{ 2 } }$$$$=\sqrt { 500 } =10\sqrt { 5 }$$After explosion $$m$$ is get $$\cfrac { 1 }{ 2 } m$$, So, by conservation of energy velocity will be double$$=20\sqrt { 5 }$$Height travelled after explosion$$=\cfrac { u^{ 2 }\sin ^{ 2 }{ \theta } }{ 2g }$$$$=\cfrac { (20\sqrt { 5 } )^{ 2 }\times \sin ^{ 2 }{ \theta } }{ 2g }$$$$=\cfrac { 400\times 5\times \cfrac { 1 }{ 5 } }{ 2\times 10 }$$$$=20$$Total height$$=20+15=35cm$$Physics

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