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

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$\frac{1}{3} s$

$\frac{1}{2} s$

1 s

$\frac{1}{4} s$

answer is B.

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Detailed Solution

Acceleration of the ball relative to elevator,

$a_{r} = a_{b} - a_{e} = (- 10) - (+ 2) = - 12 m / s^{2}$

Now, $T = \frac{2 u_{y}}{|a_{r}|} = \frac{2 \times u \sin θ}{|a_{r}|}$

$= \frac{2 \times 4 \times \sin 30 °}{12} = \frac{1}{3} s$

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