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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
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B
50m
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C
15m
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D
35m
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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$$=20+15=35cm$$

962286_758808_ans_2442eb25868e4e0082ce4a85a8da9e24.PNG

Physics

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