**Terminal velocity** (v_{t}) 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** (C_{D}). Using the equation of terminal velocity: **v _{t} = √2 m g/ρ A C_{D}**, 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/m^{3}, the cross sectional area of a parachute bag is 0.10 m^{2}, and the drag coefficient is 0.12. (Take the value of gravitational acceleration, g = 9.81 m/s^{2})

Solution:

Given data:

Terminal velocity of a parachute bag, v_{t} = ?

Mass of a parachute bag, m = 0.5 kg

Density of the air, ρ = 1.21 kg/m^{3}

Cross sectional area of a parachute bag, A = 0.10 m^{2}

Drag coefficient, C_{D} = 0.12

Gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of terminal velocity,

v_{t} = √2 m g/ρ A C_{D}

v_{t} = √(2 × 0.5 × 9.81)/(1.21 × 0.10 × 0.12)

v_{t} = √(9.81)/(0.0145)

v_{t} = √676.5517

v_{t} = 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/m^{3}, the cross sectional area of a skydiver is 0.20 m^{2}, and the drag coefficient is 0.70.

Solution:

Given data:

Mass of a skydiver, m = 55 kg

Terminal velocity of a skydiver, v_{t} = ?

Density of the air, ρ = 1.25 kg/m^{3}

Cross sectional area of a skydiver, A = 0.20 m^{2}

Drag coefficient, C_{D} = 0.70

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of terminal velocity,

v_{t} = √2 m g/ρ A C_{D}

v_{t} = √(2 × 55 × 9.81)/(1.25 × 0.20 × 0.70)

v_{t} = √(1079.1)/(0.175)

v_{t} = √6166.2857

v_{t} = 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 “v_{t}”. If the density of the air is 1.22 kg/m^{3}, the cross sectional area of a leaf is 0.05 m^{2}, 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/m^{3}

Cross sectional area of a leaf, A = 0.05 m^{2}

Drag coefficient, C_{D} = 0.08

Terminal velocity of a leaf, v_{t} = ?

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of terminal velocity,

v_{t} = √2 m g/ρ A C_{D}

v_{t} = √(2 × 0.05 × 9.81)/(1.22 × 0.05 × 0.08)

v_{t} = √(0.981)/(0.0048)

v_{t} = √204.375

v_{t} = 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/m^{3}, the cross sectional area of a feather is 0.02 m^{2}, and the drag coefficient is 0.05. (Take the value of gravitational acceleration, g = 9.81 m/s^{2})

Solution:

Given data:

Terminal velocity of a feather, v_{t} = ?

Mass of a feather, m = 0.012 kg

Density of the air, ρ = 1.20 kg/m^{3}

Cross sectional area of a feather, A = 0.02 m^{2}

Drag coefficient, C_{D} = 0.05

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of terminal velocity,

v_{t} = √2 m g/ρ A C_{D}

v_{t} = √(2 × 0.012 × 9.81)/(1.20 × 0.02 × 0.05)

v_{t} = √(0.2354)/(0.0012)

v_{t} = √196.1666

v_{t} = 14 m/s

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

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