Please tell the process

Question

When a balloon rising vertically upwards at a velocity of 10 m/* is at a height of 45 m from the ground a parachutist bails out from the balloon. After 3 seconds he opens the parachute and decelerates at a constant rate of 5 m/s2.(a)What is the height of the parachutist above the ground when he opens the parachute?(b)how far is he from the balloon at this instant?(c)with what velocity does he strike the ground(d)what time does he take in striking the ground after his exist from the balloon?

Answer 1

Lets find the displacement in the 1st 3 seconds after the parachutist bails out. We will consider any thing with direction away from the ground as positive and any thing towards the ground as negative.

This means that after 3s the parachutist is 14. 1 m below from the point where he left the balloon. So height above the ground would be, h = 45 − 14. 1

h = 30. 9 m