# Terminal Velocity Equation | Problems (With Answers)

Terminal velocity (vt) is equal to the square root of 2 times the product of mass (m) and gravitational acceleration (g) divided by the product of density (ρ), cross sectional area (A) and the drag coefficient (CD). Using the equation of terminal velocity: vt = √2 m g/ρ A CD, the value of terminal velocity of a falling object can be calculated.

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Terminal Velocity Practice Problems

Problem 1: What is the value of terminal velocity of a 0.5 kg parachute bag thrown from a plane? The density of the air is 1.21 kg/m3, the cross sectional area of a parachute bag is 0.10 m2, and the drag coefficient is 0.12. (Take the value of gravitational acceleration, 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

Using the equation of terminal velocity,
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 55 kg skydiver jumps from a helicopter, he achieves the terminal velocity after travelling some distance. Calculate the terminal velocity of a skydiver, if the density of the air is 1.25 kg/m3, the cross sectional area of a skydiver is 0.20 m2, and the drag coefficient is 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
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of terminal velocity,
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: One leaf of mass 0.05 kg is floating in the air and falling down from some height achieves the terminal velocity “vt”. If the density of the air is 1.22 kg/m3, the cross sectional area of a leaf is 0.05 m2, and the drag coefficient is 0.08, then find the value of terminal velocity of a leaf.

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 = ?
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of terminal velocity,
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 which is floating in the air. If the density of the air is 1.20 kg/m3, the cross sectional area of a feather is 0.02 m2, and the drag coefficient is 0.05. (Take the value of gravitational acceleration, 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
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of terminal velocity,
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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