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# A very broad elevator is going up vertically with a constant acceleration of $2m/{s^2}$ . 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 ${30^0}$ . The time taken by the ball to return the floor is ($g = 10m/{s^2}$)(A) $\dfrac{1}{2}s$(B) $\dfrac{1}{3}s$(C) $\dfrac{1}{4}s$(D) $1s$

Last updated date: 10th Sep 2024
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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
$T = \dfrac{{2v\sin \theta }}{{{a_{net}}}}$

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 ${v_y} = 4\sin {30^0} = 2m/s$
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 ${30^0}$