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)
h = ut + ½ at2
=> -98 = 14t - 0.5 × 9.8 × t2
=> t = 6 s
Thus, the packet reaches the ground in 6 s.
v = u + at
=> v = 14 – 9.8 × 6 = -44.8 m/s (downward)
This is the velocity with which the packet reaches the ground.