On a rainy day, a raindrop falls from very high clouds and faces retardation due to air. This retardation is directly proportional to the instantaneous speed of the drop and proportionality constant is given as $α$ . An expression for the distance travelled $s$ by the drop in time $t$ is: (given acceleration due to gravity $= g$ )

s = \frac{g}{α^{2}} (e^{- α t} - 1) + \frac{g}{α} t

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s = \frac{g t}{α}

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s = \frac{g}{α^{2}} (e^{- α t} - 1)

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s = \frac{g t}{α^{2}} + \frac{g}{α} (e^{- α t} - 1)

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Solution

The correct option is A $s = \frac{g}{α^{2}} (e^{- α t} - 1) + \frac{g}{α} t$
Let the retardation due to air be $α V$
Total acceleration of the drop $a = g - α V$
$\frac{d V}{d t} = g - α V$
Integrating from $0$ to $t$ ,
$- \frac{1}{α} l n (\frac{g - α V}{g} = t$
$V = \frac{g - g e^{- α t}}{α}$
$\frac{d S}{d t} = \frac{g - g e^{- α t}}{α}$
Integrating from $0$ to $t$ ,
$S = \frac{g t}{α} + (\frac{g}{α^{2}}) (e^{- α t} - 1)$

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The correct option is A s=gα2(e−αt−1)+gαtLet the retardation due to air be αVTotal acceleration of the drop a=g−αVdVdt=g−αVIntegrating from 0 to t , −1αln(g−αVg=tV=g−ge−αtα dSdt=g−ge−αtα Integrating from 0 to t , S=gtα+(gα2)(e−αt−1)