(a) A small mail bag is released from a helicopter

that is descending steadily at 2.13 m/s.
After 4.38 s, what is the speed of the mail-
bag? The acceleration of gravity is 9.8 m/s2 .

(b) After the mailbag is dropped, the helicopter
continues descending for 1 s but then stops.
How far is the mailbag below the helicopter
at 4.38 s?

(c) What would be the speed of the mailbag if
the helicopter had been rising steadily at u =
2.13 m/s ? (Take down as positive.)
Answer in units of m/s

(a) To find the speed of the mailbag after 4.38 seconds, we can use the formula for velocity:

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

In this case, the initial velocity u is 0 m/s (since the mailbag is released from a helicopter in descent), the acceleration a is the acceleration due to gravity -9.8 m/s^2 (acting downward), and the time t is 4.38 seconds.

Plugging in the values into the formula, we get:

v = 0 + (-9.8) * 4.38

v = -42.84 m/s

Therefore, the speed of the mailbag after 4.38 seconds is 42.84 m/s downward (negative due to the direction of acceleration).

(b) Since the helicopter continues descending for 1 second after the mailbag is dropped, the distance traveled by the mailbag during this additional time can be calculated using the formula for distance:

s = ut + (1/2)at^2

Where:
s = distance
u = initial velocity
a = acceleration
t = time

In this case, the initial velocity u is 0 m/s, the acceleration a is -9.8 m/s^2 (acting downward), and the time t is 4.38 seconds.

Plugging in the values, we get:

s = 0 * 4.38 + (1/2) * (-9.8) * (4.38)^2

s = -45.96 m

Therefore, the mailbag is 45.96 meters below the helicopter at 4.38 seconds.

(c) If the helicopter had been rising steadily at a velocity of 2.13 m/s, we can calculate the speed of the mailbag using the same formula as in part (a):

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

In this case, the initial velocity u is 2.13 m/s (upward), the acceleration a is -9.8 m/s^2 (acting downward due to gravity), and the time t is 4.38 seconds.

Plugging in the values, we get:

v = 2.13 + (-9.8) * 4.38

v = -41.81 m/s

Therefore, the speed of the mailbag would be 41.81 m/s downward (negative) if the helicopter had been rising steadily at a velocity of 2.13 m/s.

(a) To find the speed of the mailbag after 4.38 s, we need to calculate its velocity.

Given:
Initial velocity (u) = 0 m/s (since the mailbag is released from rest)
Acceleration (a) = acceleration due to gravity = 9.8 m/s^2
Time (t) = 4.38 s

The formula to calculate velocity (v) is:
v = u + a * t

Substituting the given values:
v = 0 + 9.8 * 4.38

Calculating:
v = 42.924 m/s

Therefore, the speed of the mailbag after 4.38 s is 42.924 m/s.

(b) To find how far the mailbag is below the helicopter at 4.38 s, we can use the equation of motion:

y = ut + 1/2 * a * t^2

Given:
Initial velocity (u) = 0 m/s (since the mailbag is released from rest)
Acceleration (a) = acceleration due to gravity = 9.8 m/s^2
Time (t) = 4.38 s

Substituting the given values:
y = 0 + 1/2 * 9.8 * (4.38)^2

Calculating:
y = 90.7932 m

Therefore, the mailbag is 90.7932 m below the helicopter at 4.38 s.

(c) If the helicopter had been rising steadily at a velocity u = 2.13 m/s, the speed of the mailbag would be the vector sum of the helicopter's velocity and the acceleration due to gravity.
Since the direction of the helicopter's velocity is upward, we will consider it as positive.

Given:
Helicopter velocity (u) = 2.13 m/s (upward)
Acceleration due to gravity (a) = -9.8 m/s^2 (downward)

The resulting velocity (v) is given by:
v = u + a

Substituting the given values:
v = 2.13 + (-9.8)

Calculating:
v = -7.67 m/s

Therefore, the speed of the mailbag if the helicopter had been rising steadily at 2.13 m/s would be 7.67 m/s downward.

Part a)

Assume that downward is positive for this problem.

V=V_o + g*t

V_o = 2.13
g=9.8
t=4.38

V=2.13 + (9.8*4.38)
V = 45.1 m/s

part b)

x= V_o*t + (1/2)*g*t^2
x=(2.13*4.38)+(1/2)*9.81*4.38^2
x=103.43

note that the helicopter is going down at:
x1=2.13*4.38 = 9.33

X = x - x1
X = 103.43 - 9.33
X = 94.1 m

part c)
question states that assume down is positive, and that the helicopter is raising so:

V_o = -2.13

V=V_o + g*t
V= -2.13+(9.8*4.38)
V = 40.79 m/s