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Question-169317

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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?
$${find}\:{time}\:{of}\:{flight}\:{also}? \\ $$
Commented by mr W last updated on 29/Apr/22
consider air resistance? then i think you can′t solve the problem.
$${consider}\:{air}\:{resistance}? \\ $$$${then}\:{i}\:{think}\:{you}\:{can}'{t}\:{solve}\:{the}\:{problem}. \\ $$
Commented by BHOOPENDRA last updated on 29/Apr/22
yes
$${yes} \\ $$
Answered by mahdipoor last updated on 28/Apr/22
V(t)=(−v+v_0 cosθ , v_0 sinθ−gt) get (v_0 cosθ−v=a) and (v_0 sinθ=b) p is meaning point of rocket V=(dp/dt) ⇒ p(t)=(at+x_0 ,bt−((gt^2 )/2)+y_0 ) ⇒p(0)=(0,l) ⇒ x_0 =0 , y_0 =l p(t)=(at,bt−((gt^2 )/2)+l)
$$\mathrm{V}\left({t}\right)=\left(−{v}+{v}_{\mathrm{0}} {cos}\theta\:,\:{v}_{\mathrm{0}} {sin}\theta−{gt}\right) \\ $$$${get}\:\left({v}_{\mathrm{0}} {cos}\theta−{v}={a}\right)\:{and}\:\left({v}_{\mathrm{0}} {sin}\theta={b}\right) \\ $$$${p}\:{is}\:{meaning}\:\:{point}\:{of}\:{rocket} \\ $$$$\mathrm{V}=\frac{{dp}}{{dt}}\:\Rightarrow\:{p}\left({t}\right)=\left({at}+{x}_{\mathrm{0}} ,{bt}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}+{y}_{\mathrm{0}} \right) \\ $$$$\Rightarrow{p}\left(\mathrm{0}\right)=\left(\mathrm{0},{l}\right)\:\Rightarrow\:{x}_{\mathrm{0}} =\mathrm{0}\:,\:{y}_{\mathrm{0}} ={l} \\ $$$${p}\left({t}\right)=\left({at},{bt}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}+{l}\right) \\ $$
Commented by BHOOPENDRA last updated on 28/Apr/22
wonderful sir but you are lacking 2,3 things...Let me wait for mr.W sir solution.
$${wonderful}\:{sir}\:{but}\:{you}\:{are}\:{lacking}\:\mathrm{2},\mathrm{3} \\ $$$${things}…{Let}\:{me}\:{wait}\:{for}\:{mr}.{W}\:{sir}\: \\ $$$${solution}. \\ $$
Commented by mr W last updated on 29/Apr/22
from mahdipoor sir′s result you can get answers to all other quedtions
$${from}\:{mahdipoor}\:{sir}'{s}\:{result}\:{you}\:{can} \\ $$$${get}\:{answers}\:{to}\:{all}\:{other}\:{quedtions} \\ $$
Answered by mr W last updated on 30/Apr/22
V_x =v_0 cos θ−v V_y =v_0 sin θ−gt x(t)=V_x t=(v_0 cos θ−v)t y(t)=l+v_0 sin θ t−((gt^2 )/2) at y(t)=l+h: V_y =0 v_0 sin θ−gt=0 ⇒t=((v_0 sin θ)/g) l+h=y(t)_(max) =l+(v_0 sin θ−(g/2)×((v_0 sin θ)/g))((v_0 sin θ)/g) ⇒h=((v_0 ^2 sin^2 θ)/(2g)) at x(t)=X: y(t)=0 l+v_0 sin θ t−((gt^2 )/2)=0 ⇒t=((v_0 sin θ+(√(v_0 ^2 sin^2 θ+2gl)))/g) X=x(t)=(((v_0 cos θ−v)(v_0 sin θ+(√(v_0 ^2 sin^2 θ+2gl))))/g) R=(√(X^2 +l^2 )) total flight time T=((v_0 sin θ+(√(v_0 ^2 sin^2 θ+2gl)))/g) total distance traveled: s=∫_0 ^T V dt=∫_0 ^T (√(V_x ^2 +V_y ^2 )) dt s=∫_0 ^T (√((v_0 cos θ−v)^2 +(v_0 sin θ−gt)^2 ))dt s=−(1/g)∫_0 ^T (√((v_0 cos θ−v)^2 +(v_0 sin θ−gt)^2 ))d(v_0 sin θ−gt) s=(1/g)∫_(−(√(v_0 ^2 sin^2 θ+2gl))) ^(v_0 sin θ) (√((v_0 cos θ−v)^2 +u^2 ))du ......
$${V}_{{x}} ={v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v} \\ $$$${V}_{{y}} ={v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt} \\ $$$${x}\left({t}\right)={V}_{{x}} {t}=\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right){t} \\ $$$${y}\left({t}\right)={l}+{v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${at}\:{y}\left({t}\right)={l}+{h}:\:{V}_{{y}} =\mathrm{0} \\ $$$${v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}=\mathrm{0}\:\Rightarrow{t}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}} \\ $$$${l}+{h}={y}\left({t}\right)_{{max}} ={l}+\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−\frac{{g}}{\mathrm{2}}×\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\right)\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}} \\ $$$$\Rightarrow{h}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{g}} \\ $$$$ \\ $$$${at}\:{x}\left({t}\right)={X}:\:{y}\left({t}\right)=\mathrm{0} \\ $$$${l}+{v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0}\: \\ $$$$\Rightarrow{t}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta+\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}}{{g}} \\ $$$${X}={x}\left({t}\right)=\frac{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta+\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}\right)}{{g}} \\ $$$$ \\ $$$${R}=\sqrt{{X}^{\mathrm{2}} +{l}^{\mathrm{2}} } \\ $$$$ \\ $$$${total}\:{flight}\:{time}\:{T}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta+\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}}{{g}} \\ $$$$ \\ $$$${total}\:{distance}\:{traveled}: \\ $$$${s}=\int_{\mathrm{0}} ^{{T}} {V}\:{dt}=\int_{\mathrm{0}} ^{{T}} \sqrt{{V}_{{x}} ^{\mathrm{2}} +{V}_{{y}} ^{\mathrm{2}} }\:{dt} \\ $$$${s}=\int_{\mathrm{0}} ^{{T}} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }{dt} \\ $$$${s}=−\frac{\mathrm{1}}{{g}}\int_{\mathrm{0}} ^{{T}} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }{d}\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right) \\ $$$${s}=\frac{\mathrm{1}}{{g}}\int_{−\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gl}}} ^{{v}_{\mathrm{0}} \mathrm{sin}\:\theta} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta−{v}\right)^{\mathrm{2}} +{u}^{\mathrm{2}} }{du} \\ $$$$…… \\ $$
Commented by BHOOPENDRA last updated on 30/Apr/22
well done sir
$${well}\:{done}\:{sir} \\ $$

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