A ballon is ascending at the rate of 14m/s at a height of 98m above the ground when a packet is dropped from the ballon.After how much tym and with wat velocity does it reach the ground?

For the balloon,

Initial velocity, u = 14 m/s (upward)

Acceleration, a = -9.8 m/s² (downward)

Displacement, h = -98 m (downward)

Using,

h = ut + ½ at²

=> -98 = 14t - 0.5 × 9.8 × t²

=> t = 6 s

Thus, the packet reaches the ground in 6 s.

Now, using,

v = u + at

=> v = 14 – 9.8 × 6 = -44.8 m/s (downward)

This is the velocity with which the packet reaches the ground.