Terminal Velocity

Terminal velocity is the constant speed at which an object falls through a fluid (such as air or water) when the resistance of the fluid to the object’s motion is equal to the force of gravity acting on the object.

Factors Affecting Terminal Velocity

The terminal velocity of an object depends on several factors, including:

• Mass: The greater the mass of an object, the greater its terminal velocity. This is because a more massive object experiences a greater force of gravity.
• Cross-sectional area: The larger the cross-sectional area of an object, the greater its terminal velocity. This is because a larger cross-sectional area experiences more resistance from the fluid.
• Density of the fluid: The denser the fluid, the greater the terminal velocity of an object. This is because a denser fluid exerts more resistance on the object.
• Coefficient of drag: The coefficient of drag is a measure of the resistance of an object to motion through a fluid. The higher the coefficient of drag, the greater the terminal velocity of an object.

Applications of Terminal Velocity

Terminal velocity has several applications, including:

• Parachuting: Parachutes are designed to slow down the descent of a person or object by increasing the drag force. This allows the person or object to reach a safe terminal velocity.
• Skydiving: Skydivers use their bodies to create drag and reach a terminal velocity of about 120 miles per hour (193 kilometers per hour).
• Ballistics: The terminal velocity of a bullet is an important factor in determining its range and accuracy.
• Automotive engineering: The terminal velocity of a car is an important factor in determining its fuel efficiency and safety.

Terminal velocity is a fundamental concept in physics that has a wide range of applications in everyday life. By understanding the factors that affect terminal velocity, we can design objects and systems that can safely and efficiently move through fluids.

Terminal Velocity Formula

The terminal velocity of an object is the maximum velocity it can reach when falling through a fluid (such as air or water). It is reached when the drag force acting on the object is equal to the weight of the object.

Formula

The terminal velocity of an object can be calculated using the following formula:

$$ Vt = \sqrt{(2mg/ρAC_d)} $$

where:

• $Vt$ is the terminal velocity in meters per second (m/s)
• $m$ is the mass of the object in kilograms (kg)
• $g$ is the acceleration due to gravity (9.8 m/s²)
• $\rho$ is the density of the fluid in kilograms per cubic meter (kg/m³)
• $A$ is the cross-sectional area of the object in square meters (m²)
• $Cd$ is the drag coefficient of the object

Drag Coefficient

The drag coefficient is a dimensionless number that represents the amount of drag that an object experiences when moving through a fluid. It depends on the shape of the object, its surface roughness, and the Reynolds number.

The Reynolds number is a dimensionless number that represents the ratio of the inertial forces to the viscous forces acting on an object in a fluid. It is defined as:

$$ Re = ρVD/μ $$

where:

• $\rho$ is the density of the fluid in kilograms per cubic meter (kg/m³)
• $V$ is the velocity of the object in meters per second (m/s)
• $D$ is the characteristic length of the object in meters (m)
• $\mu$ is the dynamic viscosity of the fluid in newton-seconds per square meter (N·s/m²)

Example

A skydiver with a mass of 75 kg and a cross-sectional area of 0.5 m² is falling through air with a density of 1.2 kg/m³. The drag coefficient of the skydiver is 0.7.

The terminal velocity of the skydiver can be calculated using the following formula:

$ Vt = \sqrt{(2mg/ρAC_d)} $

$ Vt = \sqrt{(2(75 kg)(9.8 m/s²)/(1.2 kg/m³)(0.5 m²)(0.7))} $

$ Vt = 56.4 m/s $

Therefore, the terminal velocity of the skydiver is 56.4 m/s.

Terminal Velocity Derivation

Terminal velocity is the constant speed at which an object falls through a fluid (such as air or water) when the resistance of the fluid is equal to the weight of the object. In this derivation, we will calculate the terminal velocity of a spherical object falling through a fluid.

Assumptions

We will make the following assumptions:

• The object is spherical.
• The fluid is incompressible and has a constant density.
• The flow of the fluid around the object is laminar.
• The object is falling at a constant speed.

Derivation

The forces acting on the object are:

Weight: The weight of the object is given by:

$$W = mg$$

where:

• $W$ is the weight of the object in newtons (N)
• $m$ is the mass of the object in kilograms (kg)
• $g$ is the acceleration due to gravity in meters per second squared (m/s²)

Drag: The drag force is the resistance of the fluid to the motion of the object. For a spherical object, the drag force is given by:

$$D = \frac{1}{2}C_D\rho Av^2$$

where:

• $D$ is the drag force in newtons (N)
• $C_D$ is the drag coefficient
• $\rho$ is the density of the fluid in kilograms per cubic meter (kg/m³)
• $A$ is the cross-sectional area of the object in square meters (m²)
• $v$ is the velocity of the object in meters per second (m/s)

At terminal velocity, the weight of the object is equal to the drag force:

$$W = D$$

Substituting the expressions for $W$ and $D$, we get:

$$mg = \frac{1}{2}C_D\rho Av^2$$

Solving for $v$, we get:

$$v = \sqrt{\frac{2mg}{C_D\rho A}}$$

This is the equation for the terminal velocity of a spherical object falling through a fluid.

Example

Let’s calculate the terminal velocity of a steel ball with a diameter of 1 cm falling through water. The density of steel is 7850 kg/m³ and the density of water is 1000 kg/m³. The drag coefficient for a sphere is approximately 0.5.

Substituting these values into the equation for terminal velocity, we get:

$$v = \sqrt{\frac{2(7850 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(10^{-2} \text{ m})^3}{0.5(1000 \text{ kg/m}^3)(\pi (10^{-2} \text{ m})^2)}} = 0.98 \text{ m/s}$$

Therefore, the terminal velocity of the steel ball is 0.98 m/s.

Terminal Velocity Examples

Terminal velocity is the constant speed at which an object falls through a fluid (usually air) due to the balance between the force of gravity pulling the object down and the resistance of the fluid pushing the object up. Here are some examples of terminal velocities for different objects in air:

Skydiver

• A skydiver in a spread-eagle position has a terminal velocity of approximately 120 miles per hour (193 kilometers per hour).
• A skydiver in a streamlined position has a terminal velocity of approximately 175 miles per hour (282 kilometers per hour).

Raindrop

• A small raindrop with a diameter of 1 millimeter has a terminal velocity of approximately 10 miles per hour (16 kilometers per hour).
• A large raindrop with a diameter of 5 millimeters has a terminal velocity of approximately 20 miles per hour (32 kilometers per hour).

Snowflakes

• A small snowflake with a diameter of 1 millimeter has a terminal velocity of approximately 1 mile per hour (1.6 kilometers per hour).
• A large snowflake with a diameter of 5 millimeters has a terminal velocity of approximately 5 miles per hour (8 kilometers per hour).

Dust Particles

• A small dust particle with a diameter of 1 micrometer has a terminal velocity of approximately 0.001 miles per hour (0.0016 kilometers per hour).
• A large dust particle with a diameter of 10 micrometers has a terminal velocity of approximately 0.01 miles per hour (0.016 kilometers per hour).

Meteoroids

• A small meteoroid with a diameter of 1 meter has a terminal velocity of approximately 10 miles per hour (16 kilometers per hour).
• A large meteoroid with a diameter of 10 meters has a terminal velocity of approximately 100 miles per hour (160 kilometers per hour).

The terminal velocity of an object depends on several factors, including the object’s mass, shape, and density, as well as the density of the fluid through which it is falling.

Solved Examples on Terminal Velocity

Example 1: Calculating Terminal Velocity

A skydiver with a mass of 75 kg jumps out of a plane at an altitude of 4000 m. The air density is 1.2 kg/m³. Calculate the skydiver’s terminal velocity.

Solution:

The terminal velocity of an object is given by the equation:

$ v_t = \sqrt{(2mg/ρAC)} $

where:

• v$_t$ is the terminal velocity in m/s
• m is the mass of the object in kg
• g is the acceleration due to gravity (9.8 m/s²)
• ρ is the air density in kg/m³
• A is the cross-sectional area of the object in m²
• C is the drag coefficient

In this case, the skydiver’s mass is 75 kg, the air density is 1.2 kg/m³, and the cross-sectional area of the skydiver is approximately 0.5 m². The drag coefficient for a skydiver is approximately 0.7.

Substituting these values into the equation, we get:

$ v_t = \sqrt{(2(75 kg)(9.8 m/s²)/(1.2 kg/m³)(0.5 m²)(0.7))} $

$ v_t = 56.4 m/s $

Therefore, the skydiver’s terminal velocity is 56.4 m/s.

Example 2: Calculating the Time to Reach Terminal Velocity

A ball with a mass of 0.5 kg is dropped from a height of 100 m. The air density is 1.2 kg/m³. Calculate the time it takes for the ball to reach terminal velocity.

Solution:

The time it takes for an object to reach terminal velocity is given by the equation:

$ t = (2m/ρAC)ln[(v_t - v_i)/v_t] $

where:

• t is the time in seconds
• m is the mass of the object in kg
• ρ is the air density in kg/m³
• A is the cross-sectional area of the object in m²
• C is the drag coefficient
• v$_t$ is the terminal velocity in m/s
• v$_i$ is the initial velocity in m/s

In this case, the ball’s mass is 0.5 kg, the air density is 1.2 kg/m³, and the cross-sectional area of the ball is approximately 0.01 m². The drag coefficient for a ball is approximately 0.5. The terminal velocity of the ball is approximately 10 m/s.

Substituting these values into the equation, we get:

$ t = (2(0.5 kg)/(1.2 kg/m³)(0.01 m²)(0.5))ln[(10 m/s - 0 m/s)/10 m/s] $

$ t = 1.67 s $

Therefore, it takes the ball approximately 1.67 seconds to reach terminal velocity.

Terminal Velocity FAQs

What is terminal velocity?

Terminal velocity is the constant speed that a freely falling object reaches when the resistance of the air (drag) is equal to the force of gravity pulling the object down. At this point, the object stops accelerating and continues to fall at a constant speed.

What factors affect terminal velocity?

The terminal velocity of an object depends on several factors, including:

• Mass: The more massive an object is, the greater its terminal velocity. This is because more massive objects experience a greater force of gravity pulling them down.
• Cross-sectional area: The larger the cross-sectional area of an object, the greater its drag. This is because a larger cross-sectional area means that the object has more surface area for the air to push against.
• Shape: The shape of an object also affects its drag. Objects with streamlined shapes experience less drag than objects with irregular shapes.
• Density of the fluid: The terminal velocity of an object also depends on the density of the fluid it is falling through. Objects falling through denser fluids experience greater drag and therefore have lower terminal velocities.

What is the terminal velocity of a human?

The terminal velocity of a human in Earth’s atmosphere is approximately 120 mph (193 km/h). This value can vary depending on the person’s weight, shape, and clothing.

What is the terminal velocity of a raindrop?

The terminal velocity of a raindrop is approximately 10 mph (16 km/h). This value can vary depending on the size and shape of the raindrop.

What is the terminal velocity of a skydiver?

The terminal velocity of a skydiver in Earth’s atmosphere is approximately 120 mph (193 km/h). This value can vary depending on the skydiver’s weight, shape, and clothing.

What is the terminal velocity of a spacecraft?

The terminal velocity of a spacecraft in Earth’s atmosphere is approximately 17,500 mph (28,000 km/h). This value can vary depending on the spacecraft’s mass, shape, and the density of the atmosphere.

What is the terminal velocity of a meteor?

The terminal velocity of a meteor in Earth’s atmosphere is approximately 40,000 mph (64,000 km/h). This value can vary depending on the meteor’s mass, shape, and the density of the atmosphere.