An open elevator is moving with an upward velocity of 0.52ms-1 with an upward acceleration of 2.4ms-2. A ball bearing then rolls off the floor with zero horizontal speed. Determine the speeds of ball bearing and elevator 2s after the ball rolls off the floor. Using v=u+at, speed of ball bearing Answer given was v=(-0.52)+(9.81)(2) I don't understand why the acceleration of ball is 9.81 and not acceleration of lift. Shouldn't the acceleration of ball be the same as lift as they are in contact?