If a body moves through a liquid or a gas then the fluid applies a force on the body which is called drag force. Direction of the drag force is always opposite to the motion of the body relative to the fluid. At low speeds of the body, drag frog `( F _(P))` is directly proportional to the speed.
`F_(D) = kv `
What K is a proportionally constant and it depends upon the dimension of the body moving in air at relatively high speeds, the drag force applied by air an the body is proportional to `v^(2)`
Where this proportionally constant K can be given by
`K_(2) rho CA`
Where `rho` is the density of air
C is another constant givig the drag property of air
A is area of cross-section of the body
Consider a case an object of mass m is released from a height h and it falls under gravity. As it's speed increases the drag force starts increasing on the object. Due to this at some instant, the object attains equilibrium. The speed attained by the body at this instant is called "terminal speed" of the body.
Assume that the drag force applied by air on the body follows the relation `F_(D) = kv`,neglect the force by buoyancy applied by air on the body then answer the following questions.
What is the terminal speed of the object ?

Options

A. `( mg)/( K)`
B. `mgK`
C. `(k)/(mg)`
D. `sqrt((mg)/( K))`

Answer

To find the terminal speed of an object falling under gravity while experiencing drag force, we can follow these steps:

### Step 1: Understand the Forces Acting on the Object
When the object is falling, two main forces are acting on it:
1. The gravitational force (weight) acting downward, given by \( F_g = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
2. The drag force acting upward, given by \( F_D = kv \), where \( k \) is a proportionality constant and \( v \) is the speed of the object.

### Step 2: Set Up the Equation at Terminal Velocity
At terminal velocity, the object stops accelerating, meaning the net force acting on it is zero. Therefore, the drag force equals the gravitational force:
\[ F_D = F_g \]
This can be expressed mathematically as:
\[ kv = mg \]

### Step 3: Solve for Terminal Velocity
To find the terminal velocity \( v_t \), we can rearrange the equation:
\[ v = \frac{mg}{k} \]
Thus, the terminal speed \( v_t \) of the object is given by:
\[ v_t = \frac{mg}{k} \]

### Step 4: Substitute for \( k \)
If we know that \( k \) can be expressed as \( K = \rho C A \) (where \( \rho \) is the density of air, \( C \) is a drag coefficient, and \( A \) is the cross-sectional area), we can substitute this into our equation for terminal velocity:
\[ v_t = \frac{mg}{\rho C A} \]

### Final Answer
The terminal speed of the object is:
\[ v_t = \frac{mg}{k} \]
or, if substituting for \( k \):
\[ v_t = \frac{mg}{\rho C A} \]

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