Awater drop of mass 20g falls to the ground with a velocity of 8m/s from a height of 20m. what is the air resistance force acting on the water drop?

0.164 N

Explanation:

Newton's second law: sum of the forces equals mass times acceleration.

∑F = ma

F − mg = ma

a = (F − mg) / m

In real life, air resistance is a function of velocity.  But we will assume that it is constant.

Given:

y₀ = 20 m

y = 0 m

v₀ = 0 m/s

v = -8 m/s

Find: a

v² = v₀² + 2a (y − y₀)

(-8 m/s)² = (0 m/s)² + 2a (0 m − 20 m)

a = -1.6 m/s²

Therefore:

-1.6 m/s² = (F − 0.020 kg × 9.8 m/s²) / 0.020 kg

F = 0.164 N

Round as needed.

did yall get answers for this yet