We model the drag force of the atmosphere as Fdrag=cv2, Fdrag =c v^2, where the value of c for a

step 1 This problem covers the concept of the terminal speed. So for part A, when the parachute is open

We model the drag force of the atmosphere as Fdrag=cv2,...

We model the drag force of the atmosphere as Fdrag=cv2, $F_{\text {drag }}=c v^{2}$, where the value of $c$ for a $70.0$ -kg person with a parachute is $18.0 \mathrm{~kg} / \mathrm{m}$. (a) What is the person's terminal velocity? (b) Without a parachute, the same person's terminal velocity would be about $50.0 \mathrm{~m} / \mathrm{s}$. What would be the value of the proportionality constant $c$ in that case? Example $5-7$

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Key Concepts

Quadratic Drag Model

In many situations, particularly at higher speeds or in denser fluids, the drag force is proportional to the square of the velocity, expressed as F_drag = c * v^2. This quadratic dependency reflects the increasing influence of velocity on the magnitude of the resistive force.

Equilibrium of Forces

When an object reaches terminal velocity, the gravitational force acting downward is balanced by the upward drag force, creating an equilibrium where the net force is zero. This balance is fundamental in determining the terminal speed and understanding the motion of falling bodies.

Terminal Velocity

Terminal velocity is the constant speed that an object eventually reaches when the force of gravity pulling it downward is exactly balanced by the opposing drag force. This balance results in zero net acceleration, meaning the object continues to fall at this steady speed.

Drag Force

Drag force is the resistive force that acts opposite to the direction of motion of an object moving through a fluid, such as air. It is a key concept in dynamics and is influenced by factors like the shape, size, surface texture of the object, and the properties of the fluid.

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