a packet is released from a balloon accelerating upward with acceleration a. the acceleration of packet just after the release is

							
I THINK IT IS SAME IN MAGNITUDE AS THAT OF THE BALLOON BUT WITH THE DIRECTION CHANGED.
I HOPE IAM CORRECT
PLEASE APPROVE IF IAM CORRECT.
						

							Just after the release the packet posses only the  component of the velocity aquired. Here the accleration of the balloon has no effect on the acceration of the body. When the body is released, it is under free fall, under the effect of gravitational force exerted by earth. Hence the acceleration of the body is equal to g towards the centre of earth.
						

							Since the ballon was moving upward with acceleration a, there is a force of gravity on the packet too. The balloon is moving upward bexause the reaction force R>W, weight of the packet. As long as the packet was attached with the balloon, the packet acquired the velocity of the balloon, as the same force is acting on both of them. But as soon as the packet is released, the reaction force no longer acts on the packet, it is under free fall, under the influence of only gravitational force (neglecting viscous drag). So the acceleration of the packet becomes g.You can also relate is to a lift moving upward with acceleration a. If suddenly the strong is cut, it falls freely and its acceleration is g.Thanks for asking again.If you have any further doubt, you can re-post.THANKS
						

							
Yes, it is correct. And the direction of the velocity is upwards. Towards the motion of the balloon.
THANKS
						

							hi Dhany
						

							The accleration of the packet will be `g` acting downward as no external force other then gravity working on it.it will achieve the case of free fall. If I am wrong please ..explain the right answet
						

							
Since the object is in the influence of both the accelerations initially, It's acceleration just after release will be the vector sum of acceleration due to gravity and acceleration of the balloon.
						

							acceleration due to gravity =g.Because as soon as the packet droped from ballon,it looses contact of balloon.It is under free fall now.There fore it occurs the acceleration due to gravity.
						

							
its acceleration just after release will be vector sum of acceleration due to gravity and acceleration of the balloon so the object is in the influence of both the accelerations instantly.
164
						

							
Acco. to law of Inertia . the acceleration of the pack is g downward. after releasing because it is non inertial frame w.r.t. the inertial frame of the velocity at that instant when the pack is released