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Question-154854

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Question Number 154854 by peter frank last updated on 22/Sep/21
Commented by Tawa11 last updated on 22/Sep/21
nice
$$\mathrm{nice} \\ $$
Answered by mr W last updated on 22/Sep/21
Commented by mr W last updated on 22/Sep/21
h=L tan θ h=v_1 sin θ_1 t−((gt^2 )/2) ...(i) L=v_1 cos θ_1 t ...(ii) 0=v_2 sin θ_2 t−((gt^2 )/2) ...(iii) L=v_2 cos θ_2 t ...(iv) L=v_1 cos θ_1 t=v_2 cos θ_2 t ⇒(v_1 /v_2 )=((cos θ_2 )/(cos θ_1 )) ...(I) (ii) into (i): L tan θ=L tan θ_1 −((gL^2 )/(2v_1 ^2 cos^2 θ_1 )) ⇒(tan θ_1 −tan θ)cos^2 θ_1 =((gL)/(2v_1 ^2 )) ...(II) (iv) into (iii): 0=L tan θ_2 −((gL^2 )/(2v_2 ^2 cos^2 θ_2 )) ⇒tan θ_2 cos^2 θ_2 =((gL)/(2v_2 ^2 )) ...(III) (III)/(II): ((tan θ_2 cos^2 θ_2 )/((tan θ_1 −tan θ)cos^2 θ_1 ))=((v_1 /v_2 ))^2 ((tan θ_2 cos^2 θ_2 )/((tan θ_1 −tan θ)cos^2 θ_1 ))=((cos^2 θ_2 )/(cos^2 θ_1 )) ((tan θ_2 )/(tan θ_1 −tan θ))=1 ⇒tan θ_1 −tan θ_2 =tan θ ⇒proved!
$${h}={L}\:\mathrm{tan}\:\theta \\ $$$$ \\ $$$${h}={v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} \:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({i}\right) \\ $$$${L}={v}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} \:{t}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{0}={v}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} \:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({iii}\right) \\ $$$${L}={v}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} \:{t}\:\:\:…\left({iv}\right) \\ $$$$ \\ $$$${L}={v}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} \:{t}={v}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} \:{t} \\ $$$$\Rightarrow\frac{{v}_{\mathrm{1}} }{{v}_{\mathrm{2}} }=\frac{\mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{cos}\:\theta_{\mathrm{1}} }\:\:\:…\left({I}\right) \\ $$$$ \\ $$$$\left({ii}\right)\:{into}\:\left({i}\right): \\ $$$${L}\:\mathrm{tan}\:\theta={L}\:\mathrm{tan}\:\theta_{\mathrm{1}} −\frac{{gL}^{\mathrm{2}} }{\mathrm{2}{v}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{1}} } \\ $$$$\Rightarrow\left(\mathrm{tan}\:\theta_{\mathrm{1}} −\mathrm{tan}\:\theta\right)\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{1}} =\frac{{gL}}{\mathrm{2}{v}_{\mathrm{1}} ^{\mathrm{2}} }\:\:\:…\left({II}\right) \\ $$$$ \\ $$$$\left({iv}\right)\:{into}\:\left({iii}\right): \\ $$$$\mathrm{0}={L}\:\mathrm{tan}\:\theta_{\mathrm{2}} \:−\frac{{gL}^{\mathrm{2}} }{\mathrm{2}{v}_{\mathrm{2}} ^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{2}} \:=\frac{{gL}}{\mathrm{2}{v}_{\mathrm{2}} ^{\mathrm{2}} }\:\:\:…\left({III}\right) \\ $$$$ \\ $$$$\left({III}\right)/\left({II}\right): \\ $$$$\frac{\mathrm{tan}\:\theta_{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{2}} }{\left(\mathrm{tan}\:\theta_{\mathrm{1}} −\mathrm{tan}\:\theta\right)\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\left(\frac{{v}_{\mathrm{1}} }{{v}_{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{tan}\:\theta_{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{2}} }{\left(\mathrm{tan}\:\theta_{\mathrm{1}} −\mathrm{tan}\:\theta\right)\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{1}} } \\ $$$$\frac{\mathrm{tan}\:\theta_{\mathrm{2}} }{\mathrm{tan}\:\theta_{\mathrm{1}} −\mathrm{tan}\:\theta}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{1}} −\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{tan}\:\theta\:\Rightarrow{proved}! \\ $$
Commented by peter frank last updated on 22/Sep/21
appriciate sir.thank you
$$\mathrm{appriciate}\:\mathrm{sir}.\mathrm{thank}\:\mathrm{you} \\ $$
Commented by peter frank last updated on 22/Sep/21
another simplification from h=v_1 sin θ_1 t−((gt^2 )/2) .....(i) 0=v_2 sin θ_2 t−((gt^2 )/2) ...(ii) h−0=(v_1 sin θ_1 )t−(v_2 sin θ_2 )t h=(v_1 sin θ_1 −v_2 sin θ_2 )t from v_1 =(L/(cos θ_1 t)) v_2 =(L/(cos θ_2 t)) h=((L/(cos θ_1 t)) sin θ_1 −(L/(cos θ_2 t)) sin θ_2 )t h=L(((sin θ_1 )/(cos θ_1 ))−((sin θ_2 )/(cos θ_2 ))) h=L(tan θ_(1 ) −tan θ_2 ) h=Ltan θ h=L(tan θ_(1 ) −tan θ_2 ) Ltan θ=L(tan θ_(1 ) −tan θ_2 ) tan θ=tan θ_(1 ) −tan θ_2
$$\mathrm{another}\:\mathrm{simplification} \\ $$$$\mathrm{from}\: \\ $$$${h}={v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} \:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…..\left(\mathrm{i}\right) \\ $$$$\mathrm{0}={v}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} \:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{h}−\mathrm{0}=\left(\mathrm{v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} \right)\mathrm{t}−\left(\mathrm{v}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} \right)\mathrm{t}\: \\ $$$$\mathrm{h}=\left(\mathrm{v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} −\mathrm{v}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} \right)\mathrm{t} \\ $$$$\mathrm{from}\: \\ $$$$\mathrm{v}_{\mathrm{1}} =\frac{\mathrm{L}}{\mathrm{cos}\:\theta_{\mathrm{1}} \mathrm{t}}\:\:\:\:\:\:\:\:\mathrm{v}_{\mathrm{2}} =\frac{\mathrm{L}}{\mathrm{cos}\:\theta_{\mathrm{2}} \mathrm{t}} \\ $$$$\mathrm{h}=\left(\frac{\mathrm{L}}{\mathrm{cos}\:\theta_{\mathrm{1}} \mathrm{t}}\:\:\:\mathrm{sin}\:\theta_{\mathrm{1}} −\frac{\mathrm{L}}{\mathrm{cos}\:\theta_{\mathrm{2}} \mathrm{t}}\:\:\:\mathrm{sin}\:\theta_{\mathrm{2}} \right)\mathrm{t} \\ $$$$\mathrm{h}=\mathrm{L}\left(\frac{\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{cos}\:\theta_{\mathrm{1}} }−\frac{\mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{cos}\:\theta_{\mathrm{2}} }\right) \\ $$$$\mathrm{h}=\mathrm{L}\left(\mathrm{tan}\:\theta_{\mathrm{1}\:} −\mathrm{tan}\:\theta_{\mathrm{2}} \right) \\ $$$$\mathrm{h}=\mathrm{Ltan}\:\theta \\ $$$$\mathrm{h}=\mathrm{L}\left(\mathrm{tan}\:\theta_{\mathrm{1}\:} −\mathrm{tan}\:\theta_{\mathrm{2}} \right) \\ $$$$\mathrm{Ltan}\:\theta=\mathrm{L}\left(\mathrm{tan}\:\theta_{\mathrm{1}\:} −\mathrm{tan}\:\theta_{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{tan}\:\theta_{\mathrm{1}\:} −\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 23/Sep/21
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