Question Number 188259 by alcohol last updated on 27/Feb/23
Answered by mr W last updated on 27/Feb/23
$${m}={m}_{\mathrm{0}} {e}^{{kt}} \\ $$$${kv}_{\mathrm{0}} ={g} \\ $$$$\left({a}\right) \\ $$$$\frac{{d}\left({mv}\right)}{{dt}}=−{mg} \\ $$$${v}\frac{{dm}}{{dt}}+{m}\frac{{dv}}{{dt}}=−{mg} \\ $$$${vm}_{\mathrm{0}} {ke}^{{kt}} +{m}_{\mathrm{0}} {e}^{{kt}} \frac{{dv}}{{dt}}=−{m}_{\mathrm{0}} {e}^{{kt}} {g} \\ $$$$\Rightarrow{kv}+\frac{{dv}}{{dt}}=−{g}\:\:\:\checkmark \\ $$$$\left({b}\right) \\ $$$$\frac{{dv}}{{dt}}=−\left({kv}+{g}\right) \\ $$$$\int_{{v}_{\mathrm{0}} } ^{{v}} \frac{{dv}}{{kv}+{g}}=−\int_{\mathrm{0}} ^{{t}} {dt} \\ $$$$\mathrm{ln}\:\frac{{kv}+{g}}{{kv}_{\mathrm{0}} +{g}}=−{kt} \\ $$$$\frac{{kv}+{g}}{{kv}_{\mathrm{0}} +{g}}={e}^{−{kt}} \\ $$$${at}\:{highest}\:{point}:\:{v}=\mathrm{0} \\ $$$$\frac{\mathrm{0}+{g}}{{g}+{g}}={e}^{−{kt}} \\ $$$${e}^{−{kt}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${m}={m}_{\mathrm{0}} {e}^{{kt}} ={m}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} =\mathrm{2}{m}_{\mathrm{0}} \:\:\:\checkmark \\ $$