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Question Number 94922 by peter frank last updated on 21/May/20

Answered by mr W last updated on 22/May/20

time t:  −h=V_0 sin θt−(1/2)gt^2   ⇒t=(V_0 /g)(sin θ+(√(sin^2  θ+((2gh)/V_0 ^2 ))))  range R:  R=V_0 cos θ t=(V_0 ^2 /g)cos θ(sin θ+(√(sin^2  θ+((2gh)/V_0 ^2 ))))  f(θ)=cos θ(sin θ+(√(sin^2  θ+λ)))  with λ=((2gh)/V_0 ^2 )  such that R is maximum,  ((df(θ))/dθ)=0  cos θ(cos θ+((sin θ cos θ)/(√(sin^2  θ+λ))))−sin θ(sin θ+(√(sin^2  θ+λ)))=0  cos 2θ+((sin θ cos^2  θ)/(√(sin^2  θ+λ)))=sin θ (√(sin^2  θ+λ))  cos 2θ(√(sin^2  θ+λ))=(λ−cos 2θ)sin θ  cos^2  2θ(sin^2  θ+λ)=(λ^2 −2λcos 2θ+cos^2  2θ)sin^2  θ  2λcos^2  2θ=(λ^2 −2λcos 2θ)(1−cos 2θ)  (2+λ)cos 2θ=λ  cos 2θ=(λ/(λ+2))=(1/(1+(V_0 ^2 /(gh))))=((gh)/(V_0 ^2 +gh))  ⇒𝛉=(1/2)cos^(−1) ((gh)/(V_0 ^2 +gh))

timet:h=V0sinθt12gt2t=V0g(sinθ+sin2θ+2ghV02)rangeR:R=V0cosθt=V02gcosθ(sinθ+sin2θ+2ghV02)f(θ)=cosθ(sinθ+sin2θ+λ)withλ=2ghV02suchthatRismaximum,df(θ)dθ=0cosθ(cosθ+sinθcosθsin2θ+λ)sinθ(sinθ+sin2θ+λ)=0cos2θ+sinθcos2θsin2θ+λ=sinθsin2θ+λcos2θsin2θ+λ=(λcos2θ)sinθcos22θ(sin2θ+λ)=(λ22λcos2θ+cos22θ)sin2θ2λcos22θ=(λ22λcos2θ)(1cos2θ)(2+λ)cos2θ=λcos2θ=λλ+2=11+V02gh=ghV02+ghθ=12cos1ghV02+gh

Commented by peter frank last updated on 24/May/20

thank you sir

thankyousir

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