A very broad elevator is going up vertically with a constant acceleration of 2m/s2 . At the instant when its velocity is 4m/s a ball is projected from the floor of the lift with a speed of 4m/s , relative to the floor at an elevation of 300 . The time taken by the ball to return the floor is

 The component of the velocity of the ball relative to the elevator is

In the x-direction, the component of velocity is vx = 4cos300=2√3m/s

In the y-direction, the component of velocity is v= 4sin300=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 anet = g+2=10+2 = 12m/s2

The time of flight will be 2vsinθ/anet = 2 × 4 × sin300/12 = 1/3s

Here v is the velocity of the ball and θ is the angle of projection, here it is given 300

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