# Terminal velocity

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Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. As an object falls, gravity pulls it downward, causing it to accelerate. However, as its speed increases, so does the air resistance opposing its motion. Eventually, the upward force of air resistance equals the downward force of gravity, resulting in a net force of zero. At this point, the object stops accelerating and continues to fall at a constant speed, known as terminal velocity. The specific value of terminal velocity depends on factors such as the object’s mass, shape, and surface area, as well as the density of the medium it is falling through. For example, a skydiver in free fall will reach terminal velocity after a certain period, continuing to fall at that speed until they deploy their parachute.

Contents

## 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.