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Two balls $$A$$ and $$B$$ are simultaneously thrown. $$A$$ is thrown from the ground level with a velocity of $$20\ m{ s }^{ -1 }$$ in the upward direction and $$B$$ is thrown from a height of $$40\ m$$ in the downward direction with the same velocity. Where will the two balls meet?


A
15 m
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B
25 m
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C
35 m
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D
45 m
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Solution

The correct option is B $$15\ m$$
Let both the ball meet after time$$ t$$ and $$ x$$ be the distance covered by ball $$A$$.
So$$,\  x=20t-\dfrac { 1 }{ 2 } \left( 10 \right) { t }^{ 2 }...\left( 1 \right) $$
For ball $$B,\quad \left( 40-x \right) =20t+\dfrac { 1 }{ 2 } \left( 10 \right) { t }^{ 2 }...\left( 2 \right) $$ 
Adding $$ (1)\  \&\  (2) ,\quad 40=40t\quad or\quad t=1\ s$$ 
so  $$x=20-5=15\ m$$

Physics

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