Q. A very broad elevator is going up vertically with a constant acceleration of $2\ ms^{- 2}$. At the instant when its velocity is $4\ ms^{- 1}$ a ball is projected from the floor of the list with a speed of $4\ ms^{- 1}$ relative to the floor at an elevation of $30{^\circ}$. The time taken by the ball to return the floor is $(g = 10\ ms^{- 2})$
A) 1/2 s
B) 1/3 s
C) 1/4 s
D) 1 s
Explanation
(b) Components of velocity of ball relative to lift are
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$u_{x} = 4\cos{30{^\circ} = 2\sqrt{3}ms^{- 1}}$
and $u_{y} = 4\sin{30{^\circ}} = 2ms^{- 1}$
and acceleration of ball relative to lift is $12\ ms^{- 2}$ in negative $y$-direction or vertically downwards. Hence, time of flight
$T = \frac{2u_{y}}{12} = \frac{u_{y}}{6} = \frac{2}{6} = \frac{1}{3}s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.