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Question Number 188259 by alcohol last updated on 27/Feb/23

	Answered by mr W last updated on 27/Feb/23

	$m = m_{0} e^{k t}$ ${k v}_{0} = g$ $(a)$ $\frac{d (m v)}{d t} = - m g$ $v \frac{d m}{d t} + m \frac{d v}{d t} = - m g$ ${v m}_{0} {k e}^{k t} + m_{0} e^{k t} \frac{d v}{d t} = - m_{0} e^{k t} g$ $\Rightarrow k v + \frac{d v}{d t} = - g ✓$ $(b)$ $\frac{d v}{d t} = - (k v + g)$ $\int_{v_{0}}^{v} \frac{d v}{k v + g} = - \int_{0}^{t} d t$ $\ln \frac{k v + g}{{k v}_{0} + g} = - k t$ $\frac{k v + g}{{k v}_{0} + g} = e^{- k t}$ $a t h i g h e s t p o i n t : v = 0$ $\frac{0 + g}{g + g} = e^{- k t}$ $e^{- k t} = \frac{1}{2}$ $m = m_{0} e^{k t} = m_{0} {(\frac{1}{2})}^{- 1} = 2 m_{0} ✓$
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