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 the influence of gravity while experiencing drag force, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces Acting on the Object**:
- When the object is falling, two main forces are acting on it: the gravitational force (weight) and the drag force.
- The gravitational force \( F_g \) acting on the object is given by:
\[ F_g = mg \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

2. **Drag Force**:
- The drag force \( F_D \) acting on the object is given by the equation:
\[ F_D = kv \]
where \( k \) is a proportionality constant and \( v \) is the velocity of the object.

3. **Condition for Terminal Velocity**:
- At terminal velocity, the drag force equals the gravitational force. Therefore, we can set up the equation:
\[ F_D = F_g \]
Substituting the expressions for \( F_D \) and \( F_g \):
\[ kv = mg \]

4. **Solving for Terminal Velocity**:
- Rearranging the equation to solve for \( v \) (the terminal velocity):
\[ v = \frac{mg}{k} \]
- This equation gives us the terminal speed of the object in terms of its mass, the acceleration due to gravity, and the drag coefficient.

### Conclusion:
The terminal speed \( v \) of the object is given by:
\[ v = \frac{mg}{k} \]