A small relief package is dropped by a helicopter that is descending steadily at 2.0 m/s.

a) After 2.5 s, what is the speed of the relief package?

b) After 2.5 s, how is the package below the helicopter?

I think I need to use the kinematic equations.

You are correct, the kinematic equations can be used to solve these types of problems. Specifically, we can use the equations of motion for an object in free fall.

Let's break down the problem and solve it step by step:

a) After 2.5 seconds, we need to find the speed of the relief package. We know that the helicopter is descending steadily at 2.0 m/s. Assuming the relief package is dropped at the same instant the helicopter starts descending, we can treat the package as being in free fall.

To find the speed of the relief package, we can use the equation:

v = u + at

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

In this case, the initial velocity (u) of the relief package is 0 m/s since it starts from rest. The acceleration (a) due to gravity is approximately 9.8 m/s^2 (considering negligible air resistance). The time (t) is given as 2.5 seconds.

Plugging in the values into the equation, we get:

v = (0 m/s) + (9.8 m/s^2)(2.5 s)

Calculating this, we find:

v = 24.5 m/s

Therefore, the speed of the relief package after 2.5 seconds is 24.5 m/s.

b) After 2.5 seconds, we need to determine how far below the helicopter the package is. To find this, we can once again use the equation of motion for an object in free fall:

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

Where:
s = distance or height
u = initial velocity (speed)
t = time
a = acceleration due to gravity

Since in this scenario the package starts from rest, the equation simplifies to:

s = (1/2)at^2

Plugging in the values, we get:

s = (1/2)(9.8 m/s^2)(2.5 s)^2

Calculating this, we find:

s = 30.625 m

Therefore, after 2.5 seconds, the relief package is approximately 30.625 meters below the helicopter.