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)))](Q154715.png)
$${x}={d}\:\mathrm{cos}\:\phi={v}_{{i}} \mathrm{cos}\:\theta_{{i}} {t}\:\:\:...\left(\mathrm{1}\right) \\ $$$${y}={d}\:\mathrm{sin}\:\phi={v}_{{i}} \mathrm{sin}\:\theta_{{i}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\:\:...\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right): \\ $$$${t}=\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} } \\ $$$${put}\:{this}\:{into}\:\left(\mathrm{2}\right): \\ $$$${d}\:\mathrm{sin}\:\phi={v}_{{i}} \mathrm{sin}\:\theta_{{i}} ×\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} }−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\phi=\mathrm{sin}\:\theta_{{i}} ×\frac{\:\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta_{{i}} }−\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}^{\mathrm{2}} \:\theta_{{i}} } \\ $$$$\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}\:\theta_{{i}} }=\mathrm{sin}\:\theta_{{i}} \:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta_{{i}} \mathrm{sin}\:\phi \\ $$$$\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}\:\theta_{{i}} }=\mathrm{sin}\:\left(\theta_{{i}} −\phi\right) \\ $$$${d}=\frac{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} \mathrm{cos}\:\theta_{{i}} \:\mathrm{sin}\:\left(\theta_{{i}} −\phi\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$$ \\ $$$$\frac{{d}}{{d}\theta_{{i}} }\left(\frac{{gd}\:\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }\right)=\frac{{d}}{{d}\theta_{{i}} }\left[\mathrm{cos}\:\theta_{{i}} \:\mathrm{sin}\:\left(\theta_{{i}} −\phi\right)\right]=\mathrm{0} \\ $$$$−\mathrm{sin}\:\theta_{{i}} \mathrm{sin}\:\left(\theta_{{i}} −\phi\right)+\mathrm{cos}\:\theta_{{i}} \mathrm{cos}\:\left(\theta_{{i}} −\phi\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\theta_{{i}} −\phi\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\theta_{{i}} −\phi=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\theta_{{i}} =\frac{\pi}{\mathrm{4}}+\frac{\phi}{\mathrm{2}} \\ $$$${d}_{{max}} =\frac{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} \mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\phi}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\phi}{\mathrm{2}}\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{cos}\:\frac{\phi}{\mathrm{2}}−\mathrm{sin}\:\frac{\phi}{\mathrm{2}}\right)^{\mathrm{2}} }{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\phi\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\phi\right)}{{g}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\phi\right)} \\ $$$$\Rightarrow{d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} }{{g}\left(\mathrm{1}+\mathrm{sin}\:\phi\right)} \\ $$
Commented by mathdanisur last updated on 21/Sep/21
$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thankyou} \\ $$
Commented by physicstutes last updated on 21/Sep/21
$$+\mathrm{brilliant}+ \\ $$
Commented by Tawa11 last updated on 21/Sep/21
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$