A small mailbag is released from a helicopter that is descending steadily at 3.00 m/s.

(a) After 2.00 s, what is the speed of the mailbag?
v = m/s

(b) How far is it below the helicopter?
d = m

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 3.00 m/s?
v = m/s
d = m

To answer the question, we need to use the kinematic equations of motion. These equations relate the displacement, velocity, acceleration, and time of an object. In this case, we will use the equation:

v = u + at

where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time

(a) After 2.00 s, what is the speed of the mailbag?

First, we need to determine the acceleration. The problem states that the helicopter is descending steadily at 3.00 m/s, which means the acceleration is constant and equal to -3.00 m/s² (negative because it points downward).

Next, we can plug in the values into the equation to find the speed:

v = u + at
v = 0 + (-3.00 m/s²)(2.00 s)
v = -6.00 m/s

Therefore, the speed of the mailbag after 2.00 s is -6.00 m/s (negative indicates downward velocity).

(b) How far is it below the helicopter?

To find the displacement (distance below the helicopter), we need to use the equation:

d = ut + (1/2)at²

Since the initial velocity is 0 m/s (the mailbag is released from rest), the equation simplifies to:

d = (1/2)at²
d = (1/2)(-3.00 m/s²)(2.00 s)²
d = -6.00 m

Therefore, the mailbag is 6.00 meters below the helicopter after 2.00 seconds.

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 3.00 m/s?

If the helicopter is rising steadily at 3.00 m/s, both the acceleration and the initial velocity will be positive.

(a) The speed of the mailbag after 2.00 s will be:

v = u + at
v = 0 + (3.00 m/s²)(2.00 s)
v = 6.00 m/s

Therefore, the speed of the mailbag after 2.00 s is 6.00 m/s (positive indicates upward velocity).

(b) The distance below the helicopter will be the same calculation as in part (b), but with positive acceleration:

d = (1/2)at²
d = (1/2)(3.00 m/s²)(2.00 s)²
d = 6.00 m

Therefore, the mailbag is 6.00 meters below the helicopter after 2.00 seconds, even if the helicopter is rising steadily at 3.00 m/s.