Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
0 votes

A very broad elevator is going up vertically with a constant acceleration of $2 m/s^2$. At the instant when its velocity is $4m/s$ a ball is projeted from the floor of the lift with a speed of $4 m/s$ relative to the floor at an elevation of $30 ^{\circ}$. The time taken by the ball to return to the floor is $(g=10 m/s^2)$

\[(a)\;\frac{1}{2} s \quad (b)\;\frac{1}{3} s \quad (c)\;\frac{1}{4}s \quad (d)\;1 s\]

Can you answer this question?

1 Answer

0 votes
Component of velocity of ball relative to lift are
$u_x=4 \cos 30 ^{\circ}=2 \sqrt 3 m/s$
$u_y=4 \sin 30 ^{\circ}=2 m/s$
and acceleration of ball relative to lift is $a=10+2=12 m/s^2$ in negative y direction acting downwards.
Therefore Time of flight $T=\large\frac{2u_y}{a}$
$\qquad=\large\frac{2 \times 2}{12}$
Hence b is the correct answer.
answered Jul 1, 2013 by meena.p
edited Mar 21, 2014 by balaji

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