Relative to another coordinate system $S^{'}$ (denoted by single prime) moving with a vertical velocity ${\to v}_{0}$ , the equation of motion of the object becomes

m (\frac{d \to v}{d t}) = - m \to g - b \to v

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m (\frac{d \to v}{d t}) = - m \to g - b (\to v - \to v_{0})

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m [(\frac{d \to v}{d t} - \to v_{0})] = - m \to g - b (\to v + \to v_{0})

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m (\frac{d \to v}{d t}) = m (\to g - \to v_{0}) - b \to v

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Solution

The correct option is B $m (\frac{d \to v}{d t}) = - m \to g - b (\to v - \to v_{0})$
The acceleration due to gravity will be the same in both systems
$\to v +_{0} =^{'}$
$\Rightarrow \to v =^{'} -_{0}$
$\frac{d \to v}{d t} = \frac{d \to v}{d t} - \frac{d_{0}}{d t}$
$\frac{d \to v}{d t} = \frac{d^{'}}{d t}$
$\Rightarrow m \frac{d \to v}{d t} = m (\frac{d^{'}}{d t})$
$\Rightarrow m (\frac{d^{'}}{d t}) = - m g - b (\to v^{'} -_{0})$

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