A parachute of an open diameter of 4.4 m has a drag coefficient of CD=1.36 . Determine the term

step 1 A terminal velocity weight equals drag. So mass times gravity is equal to one half density times

A parachute of an open diameter of 4.4 m has a drag...

A parachute of an open diameter of $4.4 \mathrm{~m}$ has a drag coefficient of $C_{D}=1.36 .$ Determine the terminal velocity as the man parachutes downward at the air temperature of $20^{\circ} \mathrm{C}$. The total mass of the parachute and man is $100 \mathrm{~kg}$. Neglect the drag on the man.

Key Concepts

Air Density

Air density (?) plays a key role in determining the drag force. It depends on environmental conditions such as temperature and pressure. At a temperature of 20°C, standard approximations of air density are often used, though any deviations in these conditions will affect the computed drag force and the resulting terminal velocity.

Cross-Sectional Area Calculation

The effective area A, which is used in the drag force equation, is typically calculated based on the geometry of the object. For a circular parachute, the area is determined using the formula A = ? * (D/2)², where D is the diameter, providing the measure of the area that interacts with the airflow.

Equilibrium of Forces

At terminal velocity, the upward drag force equals the downward force of gravity (mg). This equilibrium condition allows one to set the drag force equal to the gravitational force to derive an expression for the terminal velocity.

Drag Force Equation

The drag force experienced by an object moving through a fluid such as air is given by the equation F_drag = ½ * ? * v² * C_D * A, where ? is the air density, v is the velocity, C_D is the drag coefficient, and A is the cross-sectional area. This equation quantifies the resistance encountered by the object as it moves.

Terminal Velocity

Terminal velocity is the constant speed that a falling object reaches when the force due to air resistance (drag) exactly balances the gravitational force acting on it. At this point, there is no net acceleration, and the object continues to fall at a steady speed.

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