Answer

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**Hint:**This is a question based on time of flight. We know that the net displacement along vertical direction is zero. The time taken by the ball to return to the floor is the time when it will be in air which is the time of flight. Hence, we will find displacement along horizontal direction to compute time of flight.

**Formula Used**

\[T = \dfrac{{2v\sin \theta }}{{{a_{net}}}}\]

**Complete step by step answer**

In this question, it is given that the elevator is going upwards, and a ball is projected from the floor of lift the ball will have projectile motion and the net acceleration acting on the ball will be the resultant of acceleration due to upward movement of the elevator and the projectile motion of the ball. Here, we need to find the time taken by the ball to return to the floor which is your time of flight.

Component of velocity of ball relative to elevator is

In the x-direction, the component of velocity is ${v_x} = 4\cos {30^0} = 2\sqrt 3 $m/s

In the y-direction, the component of velocity is ${v_y} = 4\sin {30^0} = 2m/s$

Since the elevator is going upwards then, there will a pseudo force and it is always in the direction opposite to acceleration of body and so the,

Effective acceleration will ${a_{net}} = g + 2 = 10 + 2 = 12m/{s^2}$

Time of flight will be $\dfrac{{2v\sin \theta }}{{{a_{net}}}} = \dfrac{{2 \times 4 \times \sin {{30}^0}}}{{12}} = \dfrac{1}{3}s$

Here v is the velocity of the ball and $\theta $is the angle of projection, here it is given ${30^0}$

**Hence the correct option is B.**

**Note:**It should be noted that velocity on any point on the path of its motion, that is your trajectory path which in case of projectile motion is a parabola, is in the direction of the tangent at that point. The parabolic path is the sum of the vertical movement and the horizontal movement by the body.

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