A body weighing 3 kg falls from a height of 30 m from a state of rest and at the moment

A body weighing 3 kg falls from a height of 30 m from a state of rest and at the moment of falling to the ground has a speed of 20 m / s. Determine the average force of air resistance when the body falls.

m = 3 kg
h = 30m
V0 = 0
V = 20m / s
g = 10m / s2
Fsopr-?
Two forces act on the body: the force of gravity m * g vertically downward and the force of air resistance, against the movement vertically upward. We write 2 Newton’s law m * a = m * g-Fcopr. Fcopr = m * g-m * a. Fcopr = m * (g-a). The body moves with acceleration a = V ^ 2/2 * h. a = (20m / s) ^ 2/2 * 30m = 6.67 m / s2. Fcopr = 3kg * (10m / s2-6.67m / s2) = 10 N.
Answer: air resistance force Fcopr = 10 N.