Question

A balloon is ascending at the rate of 9.8m/s at a height of 39.2m above the ground,when a food packet is dropped from the balloon.After how much time and with velocity does it reach the ground. g=9.8m/s2

Answer 1

Once released the packet would have the velocity of the balloon and acceleration due to gravity.
Apply eqn of motion
-39.2=(9.8 )t -(0.5)(9.8)t2 [t square
Solve to get t=4s
Apply v=u+ at
I.e. v=9.8 -(9.8)t
v=9.8-(9.8)×4
=-29.4m/s. (Negative sign represents downward velocity since we considered upward direction to be positive)