A bag is dropped from a hovering helicopter. The bag has fallen for 2.0 s. What is the bag's velocity? How far has it fallen?

vf=vi + gt, right?

d=1/2 g t^2, right?

thank youu :D

To determine the bag's velocity and distance fallen, we need to use the equations of motion.

First, let's calculate the bag's velocity after falling for 2.0 seconds.

The equation for velocity is:
v = u + a * t

Where:
v = final velocity
u = initial velocity (which is 0 since the bag was dropped)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (2.0 seconds)

Substituting the values, we get:
v = 0 + (-9.8) * 2.0

Calculating that, we find:
v = -19.6 m/s

Therefore, the bag's velocity after falling for 2.0 seconds is -19.6 m/s.

Next, let's calculate the distance the bag has fallen.

The equation for distance is:
s = u * t + 0.5 * a * t^2

Where:
s = distance fallen
u = initial velocity (0 m/s)
t = time (2.0 seconds)
a = acceleration due to gravity (-9.8 m/s^2)

Substituting the values, we get:
s = 0 * 2.0 + 0.5 * (-9.8) * (2.0)^2

Calculating that, we find:
s = -19.6 m

Therefore, the bag has fallen a distance of -19.6 meters.

To determine the velocity and distance fallen by the bag, we can use the equations of motion.

For the velocity (v), we can use the equation:

v = gt

where:
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (2.0 s)

Substituting the values, we get:

v = (9.8 m/s^2) * (2.0 s)
v = 19.6 m/s

Therefore, the velocity of the bag after falling for 2.0 seconds is 19.6 m/s.

To calculate the distance (d) fallen by the bag, we can use another equation:

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

Substituting the values, we get:

d = (1/2) * (9.8 m/s^2) * (2.0 s)^2
d = (1/2) * (9.8 m/s^2) * (4.0 s^2)
d = 0.5 * 9.8 m/s^2 * 4.0 s^2
d = 19.6 m/s * 4.0 s^2
d = 78.4 m

Therefore, the bag has fallen a distance of 78.4 meters.