Kyle is flying a helicopter and it is rising at 4.0 m/s when he releases a bag. After 3.0s

a) What is the bag's velocity? -25.4
b) How far has the bag fallen? -32.1
c) How far below the helicopter is the bag?

Helicopter:
a = -9.80 m/s^2
V = 4.0 m/s
t = 3.0 s
d = ?

c. the bag has fallen -32.1m, and the helicopter has risen 4*3 m. What is the difference?

12 - -32.1 = 44.1

I did not think it was this easy. I plugged the data into like four formulas trying to come up with an answer.

To find the distance the bag has fallen (d), we can use the equation:

d = V * t + (1/2) * a * t^2

Given:
V = 0 m/s (the bag's initial velocity is zero when released)
a = -9.80 m/s^2 (acceleration due to gravity, negative because it is acting downward)
t = 3.0 s (time elapsed)

d = 0 * 3.0 + (1/2) * (-9.80) * (3.0)^2
d = 0 + (-4.90) * 9.0
d = 0 - 44.1
d = -44.1 meters

Therefore, the bag has fallen 44.1 meters below the helicopter.

To find the distance the bag has fallen (d), we can use the equation:

d = (1/2) * a * t^2

where a is the acceleration due to gravity (a = -9.80 m/s^2) and t is the time it takes for the bag to fall (t = 3.0 s).

Plugging in the values, we get:

d = (1/2) * (-9.80 m/s^2) * (3.0 s)^2
= (1/2) * (-9.80 m/s^2) * 9.0 s^2
= (-4.90 m/s^2) * 9.0 s^2
= -44.1 m

So, the distance the bag has fallen is -44.1 meters.

Now, to find how far below the helicopter the bag is, we need to subtract the distance fallen from the initial height of the bag. Since the bag was released from the helicopter, the initial height is 0 meters.

The bag's height below the helicopter is:

h = (0 m) - (-44.1 m)
= 44.1 m

Therefore, the bag is 44.1 meters below the helicopter.