Question Number 49491 by Necxx last updated on 07/Dec/18
$${A}\:{rocket}\:{of}\:{mass}\:\mathrm{1000}{kg}\:{containing} \\ $$$${a}\:\:{propellant}\:{gas}\:{of}\:\mathrm{3000}{kg}\:{is}\:{to} \\ $$$${be}\:{launched}\:{vertically}.{If}\:{the}\:{fuel} \\ $$$${is}\:{consumed}\:{at}\:{a}\:{steady}\:{rate}\:{of} \\ $$$$\mathrm{60}{kg}/{s}.{Calculate}\:{the}\:{least}\:{velocith} \\ $$$${of}\:{the}\:{exhaust}\:{gases}\:{if}\:{the}\:{rocket} \\ $$$${and}\:{the}\:{content}\:{will}\:{just}\:{lift}\:{off} \\ $$$${the}\:{launching}\:{pad}\:{immediately} \\ $$$${after}\:{firing}? \\ $$
Answered by ajfour last updated on 07/Dec/18
$$\frac{{Mdv}}{{dt}}=\:\left({v}_{{exhaust}} \right)\frac{{dM}}{{dt}}−{Mg} \\ $$$${to}\:{just}\:{lift}\:{off}\:\:\:\:\frac{{dv}}{{dt}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{v}_{{exhaust}} =\:\frac{{Mg}}{\left({dM}/{dt}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{4000}×\mathrm{9}.\mathrm{8}}{\mathrm{60}}\:{m}/{s}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{40}×\mathrm{98}}{\mathrm{6}}\:=\:\frac{\mathrm{3920}}{\mathrm{6}}\:=\:\mathrm{653}.\mathrm{3}\:\frac{{m}}{{s}}\:. \\ $$
Commented by Necxx last updated on 07/Dec/18
$${please}\:{can}\:{you}\:{expantiate}\:{more}\:{on} \\ $$$${what}\:{brought}\:{about}\:{the}\:{formula} \\ $$$${you}\:{used}. \\ $$$${Thanks}\:{for}\:{helping}\:{mr}.{Ajfour} \\ $$