Question

A navy seal of mass 80 kg parachuted into an enemy harbor. At one point while he was falling, the resistive force of air exerted on him was 520 N. What can you determine about the motion?

Answer 1

Answer:

The motion of the parachute = 3.3 m/s²

Explanation:

Weight of the parachute - Resistive force of air = ma

W - Fₐ  = ma.................... Equation 1

making a the subject of formula in equation 1

a = (W- Fₐ)/m.................. Equation 2

Where W = weight of the parachute, Fₐ = resistive force of air, m = mass of the parachute, a = acceleration of the parachute

Constant: g = 9.8 m/s²

Given: Fₐ = 520 N, m = 80 kg

W = mg = 80 × 9.8 = 784 N,

Substituting these values into equation 2

a = (784-520)/80

a = 264/80

a = 3.3 m/s²

Therefore the motion of the parachute = 3.3 m/s²