Standing at the mouth of a deep well, a student throws a pebble vertically upward with a speed of 18ms−1. Exactly 2.0s later, he drops a second pebble (released from rest) from the same point. Ignore air resistance and take g=9.8ms−2 directed downward.
(i) Find the velocity and the vertical displacement of the first pebble relative to the point of projection at the instant the second pebble is released.
(ii) After the second pebble is dropped, determine the time interval before the two pebbles meet.
(iii) Calculate the distance of the meeting point below the student’s hand.