Terminal velocity equation

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Terminal velocity equation
Terminal velocity equation | Image: Learnool

The terminal velocity equation is used to calculate the maximum velocity reached by a falling object when the upward force of air resistance equals the downward force of gravity. It can be expressed as vt = √2 m g/ρ A CD, where vt represents the terminal velocity, m is the mass of the object, g is the gravitational acceleration, ρ is the density of the medium through which the object is falling, A is the cross-sectional area of the object, and CD is the drag coefficient.

Practice problems

Problem #1

What is the terminal velocity of a 0.5 kg parachute bag thrown from a plane? Given the density of the air as 1.21 kg/m3, the cross-sectional area of the parachute bag as 0.10 m2, and the drag coefficient as 0.12. Take the value of gravitational acceleration as g = 9.81 m/s2.

Solution

Given data:

  • Terminal velocity of a parachute bag, vt = ?
  • Mass of a parachute bag, m = 0.5 kg
  • Density of the air, ρ = 1.21 kg/m3
  • Cross sectional area of a parachute bag, A = 0.10 m2
  • Drag coefficient, CD = 0.12
  • Gravitational acceleration, g = 9.81 m/s2

Applying the formula:

  • vt = √2 m g/ρ A CD
  • vt = √(2 × 0.5 × 9.81)/(1.21 × 0.10 × 0.12)
  • vt = √(9.81)/(0.0145)
  • vt = √676.5517
  • vt = 26.01 m/s

Therefore, the terminal velocity of a parachute bag is 26.01 m/s.

Problem #2

When a skydiver weighing 55 kg jumps from a helicopter, he reaches the terminal velocity after traveling a certain distance. Calculate the terminal velocity of the skydiver, considering the density of the air as 1.25 kg/m3, the cross-sectional area of the skydiver as 0.20 m2, and the drag coefficient as 0.70.

Solution

Given data:

  • Mass of a skydiver, m = 55 kg
  • Terminal velocity of a skydiver, vt = ?
  • Density of the air, ρ = 1.25 kg/m3
  • Cross sectional area of a skydiver, A = 0.20 m2
  • Drag coefficient, CD = 0.70
  • Assuming gravitational acceleration, g = 9.81 m/s2

Applying the formula:

  • vt = √2 m g/ρ A CD
  • vt = √(2 × 55 × 9.81)/(1.25 × 0.20 × 0.70)
  • vt = √(1079.1)/(0.175)
  • vt = √6166.2857
  • vt = 78.52 m/s

Therefore, the terminal velocity of a skydiver is 78.52 m/s.

Problem #3

A leaf with a mass of 0.05 kg is floating in the air and descending from a certain height, eventually reaching its terminal velocity “vt.” If the density of the air is 1.22 kg/m3, the cross-sectional area of the leaf is 0.05 m2, and the drag coefficient is 0.08, determine the value of the leaf’s terminal velocity.

Solution

Given data:

  • Mass of a leaf, m = 0.05 kg
  • Density of the air, ρ = 1.22 kg/m3
  • Cross sectional area of a leaf, A = 0.05 m2
  • Drag coefficient, CD = 0.08
  • Terminal velocity of a leaf, vt = ?
  • Assuming gravitational acceleration, g = 9.81 m/s2

Applying the formula:

  • vt = √2 m g/ρ A CD
  • vt = √(2 × 0.05 × 9.81)/(1.22 × 0.05 × 0.08)
  • vt = √(0.981)/(0.0048)
  • vt = √204.375
  • vt = 14.29 m/s

Therefore, the terminal velocity of a leaf is 14.29 m/s.

Problem #4

Calculate the terminal velocity of a 0.012 kg feather floating in the air. Given the density of the air as 1.20 kg/m3, the cross-sectional area of the feather as 0.02 m2, and the drag coefficient as 0.05. Take the value of gravitational acceleration as g = 9.81 m/s2.

Solution

Given data:

  • Terminal velocity of a feather, vt = ?
  • Mass of a feather, m = 0.012 kg
  • Density of the air, ρ = 1.20 kg/m3
  • Cross sectional area of a feather, A = 0.02 m2
  • Drag coefficient, CD = 0.05
  • Gravitational acceleration, g = 9.81 m/s2

Applying the formula:

  • vt = √2 m g/ρ A CD
  • vt = √(2 × 0.012 × 9.81)/(1.20 × 0.02 × 0.05)
  • vt = √(0.2354)/(0.0012)
  • vt = √196.1666
  • vt = 14 m/s

Therefore, the terminal velocity of a feather is 14 m/s.

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Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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