Question

For high-speed motion through the air-such as the skydiver shown in Figure, 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 velocityv(t) of a falling body of massm if air resistance is proportional to the square of the instantaneous velocity.

Short answer

The differential equation for falling body is $m·\frac{dv}{dt}=mg-k{v}^2$.

Step-by-step solution

Newton’s second law

It states that the acceleration of an object depends upon two variables – the net force acting on the object and the mass of the object.

Given data

Consider the air resistance is proportional to the square of the instantaneous velocity v(t) .

Given the body of mass m falls vertically under the gravity then force is mg where g is the acceleration due to gravity

Given air resistance is proportional to the square of the instantaneous velocity v(t) which is kv².

The body is in motion then force$F=mg-k{v}^2......\left(1\right)$

Evaluation

By Newton's second law of motion, F = ma.

Consider

$$\begin{gathered} F=ma \\ ma=mg-k{v}^2(\text{from(1))} \\ m·\frac{dv}{dt}=mg-k{v}^2 \end{gathered}$$

Therefore, the differential equation for the velocity v(t) of a falling body of mass m is $m·\frac{dv}{dt}=mg-k{v}^2$.