A small mailbag is released from a helicopter that is descending steadily at 1.01 m/s.

(a) After 4.00 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 1.01 m/s?
v = m/s
d = m

(a) v=vₒ+gt = 1.01+9.8•4=40.21 m/s

(b) the distance covered by mailbag (downward)
s1= vₒt+gt²/2 =1.01•4+9.8•4²/2=82.44 m
the distance covered by the helicopter (downward)
s2=v•t =1.01•4= 4.04 m
Δs=s1-s2= 82.44 – 4.04=78.4m
(c) Upward motion of the mailbag
h1=v²/2g =1.01²/2•9.8=0.053 m.
This motion takes the time
v=vₒ-g•t1.
v=0
t1= vₒ/g=1.01/9.8 =0.1 s
Free fall during
t2=4-0.1 =3.9 s
h2 = g•t ²/2 = 9.8•3.9²/2 =74 m
Upward motion of the helicopter
s2=v•t =1.01•4= 4.04 m
Position of the mailbag
74-0.053 =73.947 m

Δs=73.947+4.04=77.987m.

What’s v and d equal then

(a) After 4.00 seconds, the speed of the mailbag is probably around 4.04 m/s. But don't worry, the mailbag won't be breaking any speed records!

(b) As for how far it is below the helicopter, let's just say the mailbag is having a "groundbreaking" experience. It's probably about 4.04 meters below the helicopter.

(c) Now, if the helicopter is rising steadily at 1.01 m/s, it's like reverse skydiving for the mailbag. Its speed will still be around 4.04 m/s, but this time it'll be 4.04 meters above the helicopter instead of below. Talk about high-flying mail!

To solve this problem, we can use the equations of motion. Let's denote the initial velocity of the mailbag as v₀ = 0 m/s (since it's released from rest), the acceleration due to gravity as g = 9.8 m/s², and the time as t = 4.00 s.

(a) After 4.00 s, what is the speed of the mailbag?
Using the equation v = v₀ + at, we can find the speed of the mailbag after 4.00 seconds.
v = 0 m/s + 9.8 m/s² * 4.00 s
v = 0 m/s + 39.2 m/s
v = 39.2 m/s

So, the speed of the mailbag after 4.00 seconds is 39.2 m/s.

(b) How far is it below the helicopter?
Using the equation d = v₀t + 1/2at², we can calculate the distance the mailbag has traveled below the helicopter after 4.00 seconds.
d = 0 m/s * 4.00 s + 1/2 * 9.8 m/s² * (4.00 s)²
d = 0 m + 1/2 * 9.8 m/s² * 16.00 s²
d = 0 m + 1/2 * 9.8 m/s² * 256.00 s²
d = 0 m + 4.9 m/s² * 256.00 s²
d = 4.9 * 256.00 m
d = 1254.40 m

So, the mailbag is approximately 1254.40 meters below the helicopter after 4.00 seconds.

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.01 m/s?
In this case, the initial velocity (v₀) becomes -1.01 m/s since the velocity of the mailbag and the helicopter are in opposite directions.

(a) After 4.00 s, what is the speed of the mailbag?
Using the equation v = v₀ + at, we can find the speed of the mailbag after 4.00 seconds.
v = -1.01 m/s + 9.8 m/s² * 4.00 s
v = -1.01 m/s + 39.2 m/s
v = 38.19 m/s

So, the speed of the mailbag after 4.00 seconds is 38.19 m/s.

(b) How far is it below the helicopter?
Using the equation d = v₀t + 1/2at², we can calculate the distance the mailbag has traveled below the helicopter after 4.00 seconds.
d = -1.01 m/s * 4.00 s + 1/2 * 9.8 m/s² * (4.00 s)²
d = -1.01 m/s * 4.00 s + 1/2 * 9.8 m/s² * 16.00 s²
d = -4.04 m + 1/2 * 9.8 m/s² * 256.00 s²
d = -4.04 m + 4.9 * 256.00 m
d = -4.04 m + 1254.40 m
d = 1,250.36 m

So, the mailbag is approximately 1,250.36 meters below the helicopter after 4.00 seconds when the helicopter is rising steadily at 1.01 m/s.

To find the answers to these questions, we need to use the following kinematic equations:

1. Velocity equation: v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. Displacement equation: d = ut + (1/2)at^2
where d is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

(a) After 4.00 s, what is the speed of the mailbag?

In this case, the initial velocity (u) is zero because the mailbag is released from rest. The acceleration (a) is the rate at which the helicopter is descending, which is given as -1.01 m/s^2 (negative because it is descending). The time (t) is given as 4.00 s.

Using the velocity equation, we can calculate the speed (v):

v = u + at
v = 0 + (-1.01 m/s^2)(4.00 s)
v = -4.04 m/s

The speed of the mailbag after 4.00 s is -4.04 m/s.

(b) How far is it below the helicopter?

Now, we need to use the displacement equation to find the distance below the helicopter. Since the mailbag is released from rest, the initial velocity (u) is still zero. Substituting the given values:

d = ut + (1/2)at^2
d = 0 + (1/2)(-1.01 m/s^2)(4.00 s)^2
d = -8.08 m

The mailbag is 8.08 meters below the helicopter.

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.01 m/s?

In this case, the acceleration (a) is positive because the helicopter is rising. The other values remain the same.

Using the velocity equation:

v = u + at
v = 0 + (1.01 m/s^2)(4.00 s)
v = 4.04 m/s

The speed of the mailbag after 4.00 s is 4.04 m/s.

Using the displacement equation:

d = ut + (1/2)at^2
d = 0 + (1/2)(1.01 m/s^2)(4.00 s)^2
d = 8.08 m

The mailbag is 8.08 meters below the helicopter.