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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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An open lift is moving upward with velocity $10 m/s$. It has an upward acceleration of $2 m/s^2$ . A ball is projected upward with velocity $20 m/s$ relative to ground. Find the time when ball meets the lift

 

\[(a)\;\frac{5}{3}\; seconds \quad (b)\;\frac{5}{2} \;seconds \quad (c)\;\frac{1}{3}\;seconds \quad(d)\;\frac{2}{3} \;seconds \]

Can you answer this question?
 
 

1 Answer

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At a time when ball again meets the lift
Distance covered by lift =Distance covered by ball
$10 t+\large\frac{1}{2}$$ \times 2 \times t^2=20 t-\large\frac{1}{2}$$ \times 10 \times t^2$
We get $t=0\; or\; t= \large\frac{5}{3}$
Hence a is the correct answer

 

answered Jul 3, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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