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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
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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 $(g=10 m/s^2)$

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

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1 Answer

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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 $T=\large\frac{2u_y}{a}$
$\qquad=\large\frac{2 \times 2}{12}$
$\qquad=\large\frac{1}{3}s$
Hence b is the correct answer.
answered Jul 1, 2013 by meena.p
edited Mar 21, 2014 by balaji
 

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