Question Number 169317 by BHOOPENDRA last updated on 28/Apr/22
Commented by BHOOPENDRA last updated on 28/Apr/22
$${find}\:{time}\:{of}\:{flight}\:{also}? \\ $$
Commented by mr W last updated on 29/Apr/22
$${consider}\:{air}\:{resistance}? \\ $$$${then}\:{i}\:{think}\:{you}\:{can}'{t}\:{solve}\:{the}\:{problem}. \\ $$
Commented by BHOOPENDRA last updated on 29/Apr/22
$${yes} \\ $$
Answered by mahdipoor last updated on 28/Apr/22
$$\mathrm{V}\left({t}\right)=\left(−{v}+{v}_{\mathrm{0}} {cos}\theta\:,\:{v}_{\mathrm{0}} {sin}\theta−{gt}\right) \\ $$$${get}\:\left({v}_{\mathrm{0}} {cos}\theta−{v}={a}\right)\:{and}\:\left({v}_{\mathrm{0}} {sin}\theta={b}\right) \\ $$$${p}\:{is}\:{meaning}\:\:{point}\:{of}\:{rocket} \\ $$$$\mathrm{V}=\frac{{dp}}{{dt}}\:\Rightarrow\:{p}\left({t}\right)=\left({at}+{x}_{\mathrm{0}} ,{bt}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}+{y}_{\mathrm{0}} \right) \\ $$$$\Rightarrow{p}\left(\mathrm{0}\right)=\left(\mathrm{0},{l}\right)\:\Rightarrow\:{x}_{\mathrm{0}} =\mathrm{0}\:,\:{y}_{\mathrm{0}} ={l} \\ $$$${p}\left({t}\right)=\left({at},{bt}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}+{l}\right) \\ $$
Commented by BHOOPENDRA last updated on 28/Apr/22
$${wonderful}\:{sir}\:{but}\:{you}\:{are}\:{lacking}\:\mathrm{2},\mathrm{3} \\ $$$${things}…{Let}\:{me}\:{wait}\:{for}\:{mr}.{W}\:{sir}\: \\ $$$${solution}. \\ $$
Commented by mr W last updated on 29/Apr/22
$${from}\:{mahdipoor}\:{sir}'{s}\:{result}\:{you}\:{can} \\ $$$${get}\:{answers}\:{to}\:{all}\:{other}\:{quedtions} \\ $$
Answered by mr W last updated on 30/Apr/22
$${V}_{{x}} ={v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v} \\ $$$${V}_{{y}} ={v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt} \\ $$$${x}\left({t}\right)={V}_{{x}} {t}=\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right){t} \\ $$$${y}\left({t}\right)={l}+{v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${at}\:{y}\left({t}\right)={l}+{h}:\:{V}_{{y}} =\mathrm{0} \\ $$$${v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}=\mathrm{0}\:\Rightarrow{t}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}} \\ $$$${l}+{h}={y}\left({t}\right)_{{max}} ={l}+\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−\frac{{g}}{\mathrm{2}}×\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\right)\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}} \\ $$$$\Rightarrow{h}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{g}} \\ $$$$ \\ $$$${at}\:{x}\left({t}\right)={X}:\:{y}\left({t}\right)=\mathrm{0} \\ $$$${l}+{v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0}\: \\ $$$$\Rightarrow{t}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta+\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}}{{g}} \\ $$$${X}={x}\left({t}\right)=\frac{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta+\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}\right)}{{g}} \\ $$$$ \\ $$$${R}=\sqrt{{X}^{\mathrm{2}} +{l}^{\mathrm{2}} } \\ $$$$ \\ $$$${total}\:{flight}\:{time}\:{T}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta+\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}}{{g}} \\ $$$$ \\ $$$${total}\:{distance}\:{traveled}: \\ $$$${s}=\int_{\mathrm{0}} ^{{T}} {V}\:{dt}=\int_{\mathrm{0}} ^{{T}} \sqrt{{V}_{{x}} ^{\mathrm{2}} +{V}_{{y}} ^{\mathrm{2}} }\:{dt} \\ $$$${s}=\int_{\mathrm{0}} ^{{T}} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }{dt} \\ $$$${s}=−\frac{\mathrm{1}}{{g}}\int_{\mathrm{0}} ^{{T}} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }{d}\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right) \\ $$$${s}=\frac{\mathrm{1}}{{g}}\int_{−\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}} ^{{v}_{\mathrm{0}} \mathrm{sin}\:\theta} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +{u}^{\mathrm{2}} }{du} \\ $$$$…… \\ $$
Commented by BHOOPENDRA last updated on 30/Apr/22
$${well}\:{done}\:{sir} \\ $$