Question

a balloon is ascending at the rate of 14 metre per second at a height of 19 m above the ground when a packet is dropped from it after how much time and with what velocity does it reach the ground.

please give answer in explaining ​

Answer 1

Answer:

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.

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Answer 2

Answer:

Explanation:

Speed =distance÷time

Time =distance ÷speed

19÷14

=1.357sec