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Question-154661




Question Number 154661 by physicstutes last updated on 20/Sep/21
Answered by mr W last updated on 20/Sep/21
x=d cos φ=v_i cos θ_i t   ...(1)  y=d sin φ=v_i sin θ_i t−(1/2)gt^2    ...(2)  from (1):  t=((d cos φ)/(v_i  cos θ_i ))  put this into (2):  d sin φ=v_i sin θ_i ×((d cos φ)/(v_i  cos θ_i ))−(1/2)g(((d cos φ)/(v_i  cos θ_i )))^2   sin φ=sin θ_i ×(( cos φ)/(cos θ_i ))−((gd)/(2v_i ^2 ))×((cos^2  φ)/(cos^2  θ_i ))  ((gd)/(2v_i ^2 ))×((cos^2  φ)/(cos θ_i ))=sin θ_i  cos φ−cos θ_i sin φ  ((gd)/(2v_i ^2 ))×((cos^2  φ)/(cos θ_i ))=sin (θ_i −φ)  d=((2v_i ^2 cos θ_i  sin (θ_i −φ))/(g cos^2  φ))    (d/dθ_i )(((gd cos^2  φ)/(2v_i ^2 )))=(d/dθ_i )[cos θ_i  sin (θ_i −φ)]=0  −sin θ_i sin (θ_i −φ)+cos θ_i cos (θ_i −φ)=0  cos (2θ_i −φ)=0  ⇒2θ_i −φ=(π/2)  ⇒θ_i =(π/4)+(φ/2)  d_(max) =((2v_i ^2 cos ((π/4)+(φ/2)) sin ((π/4)−(φ/2)))/(g cos^2  φ))  d_(max) =((v_i ^2 (cos (φ/2)−sin (φ/2))^2 )/(g cos^2  φ))  d_(max) =((v_i ^2 (1−sin φ))/(g cos^2  φ))  d_(max) =((v_i ^2 (1−sin φ))/(g(1−sin^2  φ)))  ⇒d_(max) =(v_i ^2 /(g(1+sin φ)))
$${x}={d}\:\mathrm{cos}\:\phi={v}_{{i}} \mathrm{cos}\:\theta_{{i}} {t}\:\:\:…\left(\mathrm{1}\right) \\ $$$${y}={d}\:\mathrm{sin}\:\phi={v}_{{i}} \mathrm{sin}\:\theta_{{i}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\:\:…\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right): \\ $$$${t}=\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} } \\ $$$${put}\:{this}\:{into}\:\left(\mathrm{2}\right): \\ $$$${d}\:\mathrm{sin}\:\phi={v}_{{i}} \mathrm{sin}\:\theta_{{i}} ×\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} }−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\phi=\mathrm{sin}\:\theta_{{i}} ×\frac{\:\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta_{{i}} }−\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}^{\mathrm{2}} \:\theta_{{i}} } \\ $$$$\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}\:\theta_{{i}} }=\mathrm{sin}\:\theta_{{i}} \:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta_{{i}} \mathrm{sin}\:\phi \\ $$$$\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}\:\theta_{{i}} }=\mathrm{sin}\:\left(\theta_{{i}} −\phi\right) \\ $$$${d}=\frac{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} \mathrm{cos}\:\theta_{{i}} \:\mathrm{sin}\:\left(\theta_{{i}} −\phi\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$$ \\ $$$$\frac{{d}}{{d}\theta_{{i}} }\left(\frac{{gd}\:\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }\right)=\frac{{d}}{{d}\theta_{{i}} }\left[\mathrm{cos}\:\theta_{{i}} \:\mathrm{sin}\:\left(\theta_{{i}} −\phi\right)\right]=\mathrm{0} \\ $$$$−\mathrm{sin}\:\theta_{{i}} \mathrm{sin}\:\left(\theta_{{i}} −\phi\right)+\mathrm{cos}\:\theta_{{i}} \mathrm{cos}\:\left(\theta_{{i}} −\phi\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\theta_{{i}} −\phi\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\theta_{{i}} −\phi=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\theta_{{i}} =\frac{\pi}{\mathrm{4}}+\frac{\phi}{\mathrm{2}} \\ $$$${d}_{{max}} =\frac{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} \mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\phi}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\phi}{\mathrm{2}}\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{cos}\:\frac{\phi}{\mathrm{2}}−\mathrm{sin}\:\frac{\phi}{\mathrm{2}}\right)^{\mathrm{2}} }{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\phi\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\phi\right)}{{g}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\phi\right)} \\ $$$$\Rightarrow{d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} }{{g}\left(\mathrm{1}+\mathrm{sin}\:\phi\right)} \\ $$
Commented by mathdanisur last updated on 21/Sep/21
Very nice Ser thankyou
$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thankyou} \\ $$
Commented by physicstutes last updated on 21/Sep/21
+brilliant+
$$+\mathrm{brilliant}+ \\ $$
Commented by Tawa11 last updated on 21/Sep/21
Weldone sir
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

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