Answer to Question #210632 in Differential Equations for hitch

Question

For high-speed motion through the air—such as the skydiver shown in the figure below, falling before the parachute is opened—air resistance is closer to a power of the instantaneous velocity v(t).

Determine a differential equation for the velocity v(t)

 of a falling body of mass m if air resistance is proportional to the square of the instantaneous velocity. Assume the downward direction is positive. (Use k > 0

 for the constant of proportionality, g > 0

 for acceleration due to gravity, and v for v(t).)

Answer 1

Solution

By Newton’s second law  

$m\frac{dv}{dt}=F$

Here we’ll assume v is positive when it is directed downward.

Near the Earth’s surface, the force due to gravity is mg where g is the acceleration due to gravity. Air resistance, which is proportional to the square of the instantaneous velocity, is given by -kv² where k is a positive constant. The negative sign is since the air resistance acts opposite to gravity. Thus we have the first order DE

$m\frac{dv}{dt}=mg-kv^2$