Ask Questions, Get Answers

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

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

0 votes
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

Related questions

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