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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A particle is dropped from point A at a certain height from ground. It falls freely and passes through three points B,C and D with BC=CD.Time taken for particle to move from B to C is 2 seconds and from C to D is 1 second. Time taken to move from A to B is

\[(a)\;0.5 s\quad (b)\;1.5 s \quad (c)\;0.75s \quad (d)\;0.25s\]

Can you answer this question?
 
 

1 Answer

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Let $AB=y$
$BC=CD=h$
Let time take to move from A to B=t
$y=\large\frac{1}{2} $$gt^2$-----(1)
$y+h=\large\frac{1}{2} $$g(t+2)^2$-----(2)
$y+2h=\large\frac{1}{2} $$g(t+3)^2$-----(3)
Solving we get $t=0.5 s$
Hence a is the correct answer

 

answered Jun 26, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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