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

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

Expert Answer

Part (a)

\[ \sigma F  =  ma \]

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

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

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

Where $ k $ is proportionality constant.

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

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

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 terminal velocity.

$m$ is the mass.

$g$ is the acceleration due to gravity.

The drag force exerted by an object 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 $v$ of the ball is given as:

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

-The terminal velocity 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.

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

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

(b)

The drag force exerted by an object 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 no acceleration.

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

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

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