Question

# 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
B
25 m
C
35 m
D
45 m

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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