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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. What is displacement of lift and ball at a time when ball meets the lift.

 

\[(a)\;19.4\; m \quad (b)\;9.8 m \quad (c)\;16.2 m \quad (d)\;18.6\]

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 $t=0\; or\; t= \large\frac{5}{3}$
After a time $t=5/3$ both would have been displaced by a distance
$S_L=S_b$
$\qquad=10 \times \large\frac{5}{3} +\frac{1}{2} $$\times 2 \times $$ \bigg(\large\frac{5}{3}\bigg)^2$
$\qquad=\large\frac{175}{9}$
$\qquad=19.4 m$
Hence a is the correct answer.

 

answered Jul 2, 2013 by meena.p
edited Jul 28, 2014 by thagee.vedartham
 

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