Question

a food packet is released from a helicopter which is rising steady at 2ms-. After 2 seconds,
1.What is the velocity of packet?
2.How far is it below the helicopter? ​

Answer 1

Answer:

1. 18 m/s downward direction.

2. 20 m from helicopter after 2 seconds

Explanation:

As particle is released it also get an upward velocity of 2 m/s in upward direction due to inertia.So height attained by particle due to this vilocity is

v^2 - u^ = 2gs

s = 0.2 m

now, time taken by particle to cover this 0.2 m

s  = ut + 1/2 at^2

t = 0.2 seconds

now,

time left = 2-0.2 = 1.8 seconds

so the velocity of particle after 1.8 seconds

v = u + at

v = 18 m/s in  downward direction

now,

displacement covered by particle in 1.8 seconds in downward direction

s = ut + 1/2 at^2

s= 16.2 m

so the displacement of particle from its initial after after 2 seconds is 16.2 - 16 = 16 m

Now,

the displacement covered by helicopter in 2 seconds in upward direction = 4 m

So the distance between particle and helicopter after 2 seconds = 16 m + 4 m = 20 m