A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the following.

(a) What is the speed of the package in terms of v0, g, and t? (Use any variable or symbol stated above as necessary. Let down be positive.)

(b)What distance d is it from the helicopter in terms of g and t?

(c) What are the answers in parts (a) and (b) if the helicopter is rising steadily at the same speed?

(a) If the package is dropped from a helicopter that is descending steadily at a speed v0, the initial speed of the package is also v0. The acceleration of the package is due to gravity, which is g. We can use the equation:

v = v0 + gt

where v is the speed of the package after t seconds. Therefore, the speed of the package after t seconds is:

v = v0 + gt

(b) To find the distance d, we can use the equation:

d = v0t + (1/2)gt^2

The distance d from the helicopter in terms of g and t is:

d = v0t + (1/2)gt^2

(c) If the helicopter is rising steadily at the same speed, the initial speed of the package would be -v0 (since up is now considered negative).

(a) The speed of the package after t seconds would be:

v = -v0 + gt

(b) The distance d would now be:

d = -v0t + (1/2)gt^2

(a) The speed of the package can be determined by considering the acceleration due to gravity acting on it. The package is dropped, so it experiences a downward acceleration equal to the acceleration due to gravity, denoted as g.

Since the helicopter is descending steadily at a speed v0, the initial velocity of the package is v0. As time elapses, the speed of the package increases due to the acceleration from gravity.

Using the equation of motion for uniformly accelerated motion:
v = v0 + gt

where:
v = final speed of the package
v0 = initial speed of the package
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time elapsed

Therefore, the speed of the package in terms of v0, g, and t is:
v = v0 + gt

(b) To find the distance d from the helicopter, we can use the equation of motion for uniformly accelerated motion:

d = v0t + 1/2gt^2

where:
d = distance from the helicopter
v0 = initial speed of the package
g = acceleration due to gravity
t = time elapsed

Therefore, the distance d in terms of g and t is:
d = v0t + 1/2gt^2

(c) If the helicopter is rising steadily at the same speed, the initial velocity would be upward. Let's denote the initial speed of the helicopter as v0.

(a) In this case, as the package is dropped from the helicopter, it will still experience the acceleration due to gravity acting downward, which is denoted as g. However, the initial velocity of the package would be -v0 (negative because it is in the opposite direction of the upward motion of the helicopter).

Using the same equation as in part (a):
v = v0 + gt

Substituting v0 = -v0 into the equation:
v = -v0 + gt

(b) Similarly, the distance d from the helicopter can be determined using the equation of motion:

d = v0t + 1/2gt^2

Substituting v0 = -v0 into the equation and considering the negative sign:
d = -v0t + 1/2gt^2

To answer these questions, we need to apply the laws of motion and consider the effects of gravity.

(a) The speed of the package can be determined by considering the effects of gravity. Since the helicopter is descending steadily, the initial speed v0 will decrease as time elapses. The speed of an object falling freely under gravity is given by the equation:

v = v0 - gt

Where:
v = speed of the package
v0 = initial speed of the package (provided in the question)
g = acceleration due to gravity (constant)
t = time elapsed (provided in the question)

(b) The distance, d, between the helicopter and the package can be determined by considering the motion under gravity. The distance covered by an object falling freely under gravity is given by the equation:

d = v0t - 0.5gt^2

Where:
d = distance between the helicopter and the package
v0 = initial speed of the package (provided in the question)
g = acceleration due to gravity (constant)
t = time elapsed (provided in the question)

(c) If the helicopter is rising steadily at the same speed, the equations in parts (a) and (b) remain the same. The only difference is that the direction will be opposite. In this case, the velocity of the package will be in the upward direction, while the distance will be positive.

So, to summarize:
(a) The speed of the package is given by v = v0 - gt.
(b) The distance between the helicopter and the package is given by d = v0t - 0.5gt^2.
(c) If the helicopter is rising steadily at the same speed, the equations in parts (a) and (b) remain the same, but the velocity and distance will be positive.