**(a) Write but do not solve the differential equation for the instantaneous velocity $v$ of the ball in terms of time, given quantities, quantities, and fundamental constants.**

**(b) Determine the final velocity $vt$ intervals of the given quantities and basic constants.**

The**Â article aims**Â to find the differential equation of**Â instantaneous velocity**Â and **terminal velocity**. This article uses the concept and definitions of **instantaneous and terminal velocity and related constants**.

**Expert Answer**

**Part (a)**

\[ \sigma F Â = Â ma \]

\[ w \:- \:F_{D} = ma\]

\[ mg\: -\: bv ^ { 2 } = ma \]

\[ mg\: – \: k A \delta v ^ { 2 } Â = ma \]

Where $ k $ is **proportionality constant.**

\[ a = \dfrac { dv } { dt } = Â g \:- \: (\dfrac{kA\delta}{m})v^{2} \]

\[\dfrac{dv}{dt} + \dfrac{kA \delta }{m} v^{2}= g\]

**Part (b)**

$F_{D}$ is the **drag force.**

$\delta $ is the **density**.

$A$ is the **cross-sectional area**.

$C_{D}$ is the **drag coefficient**.

$v$ is the **velocity**.

$v_{t}$ is the**Â terminal velocity**.

$m$ is the **mass**.

$g$ is the **acceleration due to gravity.**

The **drag force exerted by an object**Â when it falls from a given height is defined by the **following equation:**

\[F_{D} = \dfrac{1}{2} \delta A C_{D} v^{2}\]

Where **drag force is equal to the weight of the ball**, the terminal velocity is reached

\[mg =\dfrac{1}{2} \delta A C_{D} v_{t}^{2} \]

\[\delta A C_{D} v{t}^{2} = 2mg \]

\[v_{t} = \sqrt {\dfrac{2mg}{\delta A C_{D}}}\]

**Numerical Result**

– The **differential equation for the instantaneous speed**Â $v$ of the ball is given as:

\[\dfrac{dv}{dt} + \dfrac{kA \delta }{m} v^{2}= g\]

-The **terminal velocity**Â is given as:

\[v_{t} = \sqrt {\dfrac{2mg}{\delta A C_{D}}}\]

**Example**

A rubber ball having mass $m$ is dropped from a mountain. As the ball falls, it is subject to air drag (drag force caused by air). The drag force on ball has magnitude $av^{2}$, where $a$ is the constant drag coefficient and $v$ is instantaneous speed of the ball. The drag coefficient $a$ is directly proportional to the cross-sectional area of the ball and the air density and does not depend on the weight of the ball. As the ball falls, its velocity approaches a constant value called terminal velocity.

**(a) Write but do not solve the differential equation for the instantaneous velocity of the ball in terms of time, given quantities, quantities, and fundamental constants.**

**(b) Determine the terminal velocity $v_{t}$ intervals of the given quantities and basic constants.**

**Solution**

**(a)**

\[\sigma F = ma\]

\[w \:- \:F_{D}= ma\]

\[mg\: -\: av^{2} = ma\]

\[mg\: – \: k A \rho v^{2} = ma\]

Where $k$ is **proportionality constant.**

\[a = \dfrac{dv}{dt} = Â g \:- \: (\dfrac{kA\rho}{m})v^{2} \]

\[\dfrac{dv}{dt} + \dfrac{kA \rho }{m} v^{2}= g\]

**(b)**

The **drag force exerted by an object**Â when it falls from a given height is defined by the **following equation:**

Where **drag force is equal to the weight of the ball**, the terminal velocity is reached and there is**Â no acceleration.**

\[mg -k \rho A v_{t}^{2} = 0 \]

\[v_{t} = \sqrt {\dfrac{mg}{ k\rho A }}\]