Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

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

0 votes
during free fall
$v^2=100 g$
during deceleration
Therefore $s=293m$
Hence d is the correct answer. 


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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App