Question

A very broad elevator is going up vertically with a constant acceleration of $2ms^{- 2}$ . At the instant when its velocity is $4 \, ms^{- 1}$ a ball is projected from the floor of the elevator with a speed of $4 \, ms^{- 1}$ with respect to the elevator at an angle of $30^\circ $ . The time taken by the ball to return the floor is $\left(\right.g=10 \, ms^{- 2}\left.\right)$

• A. 1/2 s
• B. 1/3 s
• C. 1/4 s
• D. 1 s

Answer 1

Answer: (B)

Components of velocity of ball relative to lift are

$u_{x}=4cos 30 ^\circ = 2 \sqrt{3} m s^{- 1}$
and $u_{y}=4sin 30 ^\circ =2ms^{- 1}$
and acceleration of ball relative to lift is $12 \, ms^{- 2}$ in negative $y$ -direction or vertically downwards. Hence, time of flight
$T=\frac{2 u_{y}}{12}=\frac{u_{y}}{6}=\frac{2}{6}=\frac{1}{3}s$