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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at $2 m/s^2 $. He reaches the ground with a speed of 3 m/s. At what height, did he bail out $(g=10 m/s^2)$

\[(a)\;91\;m \quad (b)\;182\; m \quad (c)\;111\;m \quad (d)\;293\;m\]
Can you answer this question?
 
 

1 Answer

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during free fall
$v^2=u^2+2as$
$v^2=0+2(g)50$
$v^2=100 g$
during deceleration
$v^2=u^2+2as$
$100\;g=(3)^2+(2)(-2)s$
Therefore $s=293m$
Hence d is the correct answer. 

 

answered Jul 14, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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