Terminal velocity describes the constant speed an object eventually reaches when falling through a fluid, such as air or water. This maximum velocity occurs when the object’s downward accelerating force is perfectly balanced by the opposing force from the fluid medium. When this balance is achieved, the net force acting on the object becomes zero, and the object maintains a steady speed.
The Forces Governing Freefall
The motion of any object falling through the atmosphere is governed by two opposing forces that dictate its acceleration and eventual speed. The first is the gravitational force, which is the object’s weight, calculated as its mass multiplied by the constant acceleration due to gravity (\(F_g = mg\)). This downward force remains constant throughout the fall, continuously accelerating the object toward the Earth.
The second force is air resistance, also known as drag (\(F_d\)), which acts upward, opposing the object’s motion. Unlike gravity, the drag force starts at zero and increases as the object’s speed rises. For most falling objects, the drag force is proportional to the square of the velocity. This non-linear relationship ensures that as the object falls faster, the upward drag force quickly approaches the magnitude of the constant downward gravitational force.
Calculating the Components of Drag
To determine terminal velocity, one must first quantify the drag force using the formula \(F_d = 0.5 times rho times V^2 times C_d times A\). While velocity (\(V\)) is the variable being solved for, the other three factors—air density, cross-sectional area, and the drag coefficient—must be established.
Air density (\(rho\)) is the mass of air per unit volume and is not a fixed value, changing significantly with altitude and temperature. Density decreases at higher altitudes because the air pressure is lower, and it also decreases as air temperature rises.
The cross-sectional area (\(A\)) is the area of the object projected onto a plane perpendicular to the direction of motion, representing the “face” pushing through the air. This value depends on the object’s orientation; a skydiver belly-flopping presents a much larger area than one diving head-first, which affects the resulting drag.
The drag coefficient (\(C_d\)) is a dimensionless number determined by the object’s shape and surface texture. Streamlined shapes, like a teardrop, have a low \(C_d\) (often around 0.04), while irregular or blunt shapes, such as a box or a parachute, have a high \(C_d\) (closer to 1.0 or more).
Deriving the Terminal Velocity Equation
Determining the terminal velocity (\(V_t\)) requires finding the speed at which the two opposing forces—gravity and drag—achieve equilibrium. This state is defined by the condition where the net force on the object is zero, meaning the gravitational force (\(F_g\)) equals the drag force (\(F_d\)). The calculation begins by setting the two force equations equal: \(mg = 0.5 times rho times V_t^2 times C_d times A\).
The goal is to algebraically isolate the terminal velocity term, \(V_t\). First, multiply both sides by two to eliminate the 0.5 factor, resulting in \(2mg = rho times V_t^2 times C_d times A\). Next, move the non-velocity terms—density (\(rho\)), drag coefficient (\(C_d\)), and area (\(A\))—to the left side by dividing both sides by their product. This yields the expression for the terminal velocity squared: \(V_t^2 = frac{2mg}{rho C_d A}\).
The final step requires taking the square root of both sides to solve for \(V_t\). This produces the complete formula for calculating terminal velocity: \(V_t = sqrt{frac{2mg}{rho C_d A}}\). This equation confirms that the terminal speed is directly proportional to the square root of the object’s mass (\(m\)) and inversely proportional to the square root of the medium’s density (\(rho\)), the object’s drag coefficient (\(C_d\)), and its cross-sectional area (\(A\)).
Factors Influencing Terminal Velocity
The derived equation demonstrates how changes to the physical properties of the object or the medium directly impact terminal velocity. Objects with greater mass (\(m\)) achieve a higher terminal velocity because a larger gravitational force requires a greater balancing drag force. For two objects with the same shape and size, the heavier one must fall faster for the drag force to increase sufficiently to match its weight.
Conversely, increasing the cross-sectional area (\(A\)) or the drag coefficient (\(C_d\)) causes a reduction in terminal velocity. This is illustrated by a skydiver opening a parachute, which increases both \(A\) and \(C_d\), lowering \(V_t\) from approximately 50 meters per second (112 mph) to less than 5 meters per second (11 mph). The density of the medium (\(rho\)) also plays a role; terminal velocity is substantially lower in water than in air because water is about 800 times denser than air, creating a much larger drag force at lower speeds.