An elevator starts from rest and moves upward , accelerating at a rate of 1.2m/s^2 until it reaches a speed of 7.8m/s, which it then maintains. Two seconds after the elevator begins to move, a man standing 12m above the initial position of the top of the elevator throws a ball upward with an initial velocity of 20m/s. Determine the time when the ball will hit the elevator.

Dear Pratiek

Try to solve the problem using relative velocity and newton equations

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TIME TAKEN BY ball to reach highest point-> 0=20-9.8t =>t=2.04s

dist. covered by ball to reach the highest point->0=400-19.6s =>s=20.4m

dist covered by lift in (2+2.04)s=4.08m

distance between ball at 4.04sfrom start of lift=20.4+12-4.08=28.32

let after t seconds from falling of ball ,it hits elevator

then dist. covered by ball in t seconds=28.32m -dist covered by lift in t sec

                                          1/2gt^2=28.32m - (1/4)t^2

                                          (4.9 + 0.25)t^2=28.32

                                           5.15 t^2 =28.32

                                            t=2.34s

so ball hits elevator at (2+2.04+2.32)=6.36s