Question

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?

Answer 1

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.

Answer 2

Answer:

u=14m/s

h=-98m

a=-9.8m/s^2

now,

h=ut +1/2at^2

= -98=14t+1/2*-9.8*t^2

= t=6s

then,

v=u+at

v=14-9.8*6

=14-58.8

=-44.8 (downward)

So ,after 6s and 44.8m/s downward the food packet will reach the ground.

Explanation:

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