Terminal Velocity Formula, Definition, Concepts and Examples

Terminal Velocity Definition And Formula

Terminal velocity is the constant speed achieved by a falling object when the gravitational force pulling it downward equals the air resistance pushing against it. Initially, as an object falls, it accelerates due to gravity. However, air resistance increases with speed, eventually balancing the gravitational force. At this point, the object no longer accelerates and maintains a steady speed. Terminal velocity varies with factors like object mass, shape, and fluid density. In freefall, it is the maximum achievable speed, representing a dynamic equilibrium between gravity's pull and air resistance, ensuring the object neither speeds up nor slows down as it descends through a fluid, such as air or water.

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

Vt = √(2  (m  g) / (ρ  A  Cd))

Where:

• - `Vt` is the terminal velocity.
• - `m` represents the mass of the falling object.
• - `g` is the acceleration due to gravity (approximately 9.81 m/s² on Earth).
• - `ρ` denotes the density of the fluid (in this case, air).
• - `A` is the cross-sectional area of the object.
• - `Cd` is the drag coefficient, which is a dimensionless number representing the object's shape and how aerodynamic it is.

Detailed Descriptions

Now, let's delve deeper into each component of the terminal velocity formula:

1. Mass (m): This represents the mass of the object in kilograms. The greater the mass, the greater the gravitational force acting on it, which tends to increase the terminal velocity.
2. Acceleration due to Gravity (g): On Earth, gravity causes all objects to accelerate downward at approximately 9.81 meters per second squared. This constant value plays a crucial role in determining terminal velocity.
3. Fluid Density (ρ): The density of the fluid through which the object is falling (usually air) is essential. Denser fluids provide more resistance, leading to a lower terminal velocity.
4. Cross-sectional Area (A): The cross-sectional area of the object perpendicular to the direction of motion is a crucial factor. Objects with larger cross-sectional areas experience more air resistance, which reduces their terminal velocity.
5. Drag Coefficient (Cd): The drag coefficient is a dimensionless number that characterizes how aerodynamic an object is. A streamlined object will have a lower drag coefficient, allowing it to reach a higher terminal velocity.

How Terminal Velocity Works

When an object falls freely in a fluid, it initially accelerates due to gravity. However, as it gains speed, the air resistance opposing its motion also increases. Eventually, a point is reached where the gravitational force and the air resistance force become equal. At this moment, the object ceases to accelerate and moves at a constant speed, which is the terminal velocity.

Also Check - Relativistic Mass Formula

Examples

Example 1: Skydiver's Terminal Velocity

Scenario: Imagine a skydiver jumping out of an airplane. We want to calculate their terminal velocity during freefall.

Steps:

1. Gather Information:

• - Mass of the skydiver (m): Let's say the skydiver weighs 70 kilograms.
• - Gravity (g): On Earth, gravity is approximately 9.81 m/s².
• - Air density (ρ): At sea level, air density is roughly 1.225 kg/m³.
• - Skydiver's cross-sectional area (A): Assume the skydiver's spread-eagle position gives them an area of 0.7 m².
• - Drag coefficient (Cd): For a skydiver in a spread-eagle position, Cd is around 0.7.

2. Apply the Terminal Velocity Formula:

```

Vt = √(2  (m  g) / (ρ  A  Cd))

```

• - Plug in the values:

```

Vt = √(2  (70 kg  9.81 m/s²) / (1.225 kg/m³  0.7 m²  0.7))

```

3. Calculate:

```

Vt = √(1376.7 / (0.60225))

Vt = √2283.82

Vt ≈ 47.78 m/s

```

Explanation: In this example, we used the terminal velocity formula to calculate the skydiver's terminal velocity. With a mass of 70 kg, gravity of 9.81 m/s², air density of 1.225 kg/m³, a cross-sectional area of 0.7 m², and a drag coefficient of 0.7 for the spread-eagle position, the skydiver's terminal velocity is approximately 47.78 meters per second.

Example 2: Falling Object in Water

Scenario: Consider a spherical object, like a cannonball, falling into a swimming pool. Calculate its terminal velocity.

Steps:

1. Gather Information:

• - Mass of the object (m): Let's say the cannonball weighs 5 kilograms.
• - Gravity (g): Use the standard gravity of 9.81 m/s².
• - Water density (ρ): Water density is about 1000 kg/m³.
• - Cross-sectional area (A): For a spherical object, A is the cross-sectional area of a circle, given by πr², where r is the radius.
• - Drag coefficient (Cd): For a smooth sphere, Cd is approximately 0.47.

2. Apply the Terminal Velocity Formula:

```

Vt = √(2  (m  g) / (ρ  A  Cd))

```

• - Plug in the values:

```

Vt = √(2  (5 kg  9.81 m/s²) / (1000 kg/m³  π  (0.1 m)²  0.47))

```

3. Calculate:

```

Vt = √(98.1 / (1.479))

Vt = √66.27

Vt ≈ 8.14 m/s

```

Explanation: In this example, we calculated the terminal velocity of a 5 kg cannonball falling into water with a density of 1000 kg/m³. Using a spherical shape with a radius of 0.1 meters (giving a cross-sectional area), and a drag coefficient of 0.47 for a smooth sphere, the terminal velocity is approximately 8.14 meters per second.

Example 3: Parachute Jump

Scenario: A person is skydiving with an open parachute. Calculate their terminal velocity.

Steps:

1. Gather Information:

• - Mass of the person (m): 70 kilograms
• - Gravity (g): 9.81 m/s²
• - Air density (ρ): 1.225 kg/m³
• - Parachute's cross-sectional area (A): 44 m²
• - Drag coefficient (Cd): 1.4 (for a parachute)

2. Apply the Terminal Velocity Formula:

```

Vt = √(2  (m  g) / (ρ  A  Cd))

```

• - Plug in the values:

```

Vt = √(2  (70 kg  9.81 m/s²) / (1.225 kg/m³  44 m²  1.4))

```

3. Calculate:

```

Vt = √(1377.42 / 76.86)

Vt ≈ √17.91

Vt ≈ 4.23 m/s

```

Explanation: With a parachute, the person's terminal velocity is approximately 4.23 meters per second.

Example 4: Falling Feather

Scenario: A feather is falling through the air. Calculate its terminal velocity.

Steps:

1. Gather Information:

• - Mass of the feather (m): 0.002 kilograms
• - Gravity (g): 9.81 m/s²
• - Air density (ρ): 1.225 kg/m³
• - Feather's cross-sectional area (A): 0.00002 m²
• - Drag coefficient (Cd): 0.004 (for a feather)

2. Apply the Terminal Velocity Formula:

```

Vt = √(2  (m  g) / (ρ  A  Cd))

```

• - Plug in the values:

```

Vt = √(2  (0.002 kg  9.81 m/s²) / (1.225 kg/m³  0.00002 m²  0.004))

```

3. Calculate:

```

Vt = √(0.03924 / 0.000098)

Vt ≈ √400

Vt ≈ 20 m/s

```

Explanation: Even with a very low mass, the feather's terminal velocity is approximately 20 meters per second due to its large surface area and low drag coefficient.

Example 5: Raindrop Falling

Scenario: Calculate the terminal velocity of a raindrop falling through the air.

Steps:

1. Gather Information:

• - Mass of the raindrop (m): 0.001 kilograms
• - Gravity (g): 9.81 m/s²
• - Air density (ρ): 1.225 kg/m³
• - Raindrop's cross-sectional area (A): 0.000001 m²
• - Drag coefficient (Cd): 0.47 (typical for a spherical raindrop)

2. Apply the Terminal Velocity Formula:

```

Vt = √(2  (m  g) / (ρ  A  Cd))

```

• - Plug in the values:

```

Vt = √(2  (0.001 kg  9.81 m/s²) / (1.225 kg/m³  0.000001 m²  0.47))

```

3. Calculate:

```

Vt = √(0.01962 / 0.00000047)

Vt ≈ √41829.79

Vt ≈ 204.53 m/s

```

Explanation: Raindrops can reach terminal velocities of around 204.53 meters per second due to their small size and streamlined shape.

These examples illustrate how different objects, shapes, and sizes can lead to varying terminal velocities.

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What Are Fluids?

Fluids are substances that can flow and take the shape of their container. They can be classified into two main categories: liquids and gases.

1. Liquids:

• - Liquids are one of the two primary states of matter (the other being solids).
• - They have a definite volume but take the shape of the container they are in.
• - Liquids have a fixed density, meaning they have a specific mass per unit volume.
• - They are virtually incompressible, which means their volume doesn't change significantly under pressure.
• - Common examples of liquids include water, oil, and gasoline.

2. Gases:

• - Gases are another state of matter, characterized by their ability to fill any available space.
• - They have neither a definite shape nor a definite volume.
• - Gases are highly compressible, meaning their volume can change significantly under pressure.
• - Their density is much lower than that of liquids or solids.
• - Examples of gases include air (a mixture of gases), oxygen, and carbon dioxide.

Properties of Fluids

Fluids possess several important properties that influence their behavior:

1. Viscosity:

• - Viscosity is a measure of a fluid's resistance to flow.
• - Fluids with high viscosity are thick and flow slowly, while those with low viscosity are thin and flow quickly.
• - For example, honey has high viscosity, while water has low viscosity.

2. Density:

• - Density refers to the mass of a fluid per unit volume.
• - It determines whether a substance will float or sink in a given fluid.
• - Objects with a density lower than that of the fluid will float, while those with higher density will sink.

3. Pressure:

• - Pressure in a fluid arises from the force exerted by the fluid's molecules on a surface.
• - Pressure increases with depth in a fluid due to the weight of the fluid above.
• - This relationship is described by Pascal's principle.

4. Buoyancy:

• - Buoyancy is the upward force exerted by a fluid on an object submerged in it.
• - It counteracts the force of gravity, making objects feel lighter in a fluid.
• - Archimedes' principle explains buoyancy and why objects float.

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Fluid Dynamics

Fluid dynamics is the branch of physics that studies the motion of fluids, including how they flow, the forces that act on them, and their interactions with solid objects. Some key concepts in fluid dynamics include:

1. Bernoulli's Principle:

• - Bernoulli's principle states that as the speed of a fluid increases, its pressure decreases, and vice versa.
• - It explains phenomena like lift in aviation and the flow of fluids through pipes.

2. Flow Patterns:

• - Fluids can exhibit various flow patterns, including laminar flow (smooth and orderly) and turbulent flow (chaotic and irregular).
• - These patterns depend on factors like viscosity and velocity.

3. Reynolds Number:

• - The Reynolds number is a dimensionless quantity used to predict whether flow will be laminar or turbulent.
• - It depends on factors such as fluid density, viscosity, and the size and speed of the object.

Applications of Fluid Mechanics:

Fluid mechanics plays a vital role in various practical applications, including:

• - Designing efficient transportation systems (e.g., aircraft and cars).
• - Developing hydraulic systems for machinery.
• - Understanding weather patterns and atmospheric dynamics.
• - Designing pipelines for transporting fluids.
• - Advancing medical applications, such as blood flow in arteries.

Understanding the behavior of fluids is essential in numerous fields, from engineering and physics to environmental science and medicine, and it continues to be an area of active research and innovation.