A very broad elevator plateform is going up vertically with a constant acceleration 1 ms-2. At the instant when the velocity of the lift is 2 m/s, a stone is projected from the plateform with a speed of 20 m/s relative to the floor at an elevation 30o. The time taken by the stone to return to the floor will be

Initial velocity of stone w.r.t. lift=20sin⁡30∘j^+20cos⁡30∘i^=(103i^+10j^)m/sInitial velocity of stone w.r.t. ground=(103i^+12j^)m/sThe initial position of stone and lift are same and when they again meet their final positions will also be same. So both will have same displacement in vertical direction in same time Displacement of lift =2(t)+12×1×t2=2t+t22 Displacement of stone =12(t)−12×10×t2=12t−5t2 so  2t+t22=12t−5t211t22=10t or t=2011secSo time taken by stone to return to the floor of lift is 2011sec.