A 90 kg pilot ejects from a plane and reaches a steady state velocity of 30 m/s

Question

A 90 kg pilot ejects from a plane and reaches a steady state velocity of 30 m/s with one parachute open. Neglect any air drag force on the pilot and assume that the air drag force can be modeled as an ideal damper i.e. linearly proportional to the velocity.

A) Find the damping coefficient

An identical second parachute opens after the pilot has already reached the steady state.

B) if the 2nd parachute opens at t=0, determine the ODE describing the pilot's velocity including initial conditions.

C) Obtain the differential equation describing the force acting on the pilot.

Explanation / Answer

At steady state the weight of the pilot is equal to the breaking force of the parachute.

m*g =c*v

The damping cefficient is c.

c= (mg/v) =(90*9.8/30) =29.4 (N*s/m)

B)

The equation of motion is

mg -2*cv =m*a

The ODE for the velocity is:

m*(dv/dt) +2cv =mg

The initial condition is: at t=0 s initial spped v(0)=0 m/s

C)

The force on the pilot is

F = mg - 2cv

Here above the force on the pilot is not constant but varies with falling disatnce x (hence with time falling time t)

F(x)= mg - 2c*(dx/dt)

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