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Question Number 209359 by peter frank last updated on 08/Jul/24

Commented by mr W last updated on 08/Jul/24

“with a vertical velocity v” should be “with a velocity v”. answer ((v^2 −g^2 x^2 )/(2gv^2 )) should be ((v^4 −g^2 x^2 )/(2gv^2 )).

$$``{with}\:{a}\:{vertical}\:{velocity}\:{v}''\:{should}\:{be} \\ $$$$``{with}\:{a}\:{velocity}\:{v}''. \\ $$$${answer}\:\frac{{v}^{\mathrm{2}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} }\:{should}\:{be}\:\:\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} }. \\ $$

Commented by peter frank last updated on 08/Jul/24

true its typing arror.

$$\mathrm{true}\:\mathrm{its}\:\mathrm{typing}\:\mathrm{arror}. \\ $$

Answered by Spillover last updated on 08/Jul/24

y=xtan θ−((gx^2 )/(2u^2 ))sec^2 θ u=v let tan θ=X y=xX−((gx^2 )/(2v^2 ))(1+X^2 )......(i) (dy/dX)=x−((gx^2 )/(2v^2 ))(2X) (dy/dX)=x−((gx^2 )/(2v^2 ))(2X)=x−((gx^2 )/u^2 )X (dy/dX)=0 0=x−((gx^2 )/v^2 )X X=(v^2 /(gx)) .....(ii) y=xX−((gx^2 )/(2v^2 ))(1+X^2 ) y_(max) =x(v^2 /(gx)) −((gx^2 )/(2v^2 ))(1+(v^2 /(gx)) )=((v^4 −g^2 x^2 )/(2gv^2 )) y_(max) =((v^4 −g^2 x^2 )/(2gv^2 ))

$${y}={x}\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} }\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$${u}={v} \\ $$$${let}\:\mathrm{tan}\:\theta={X} \\ $$$${y}={xX}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+{X}^{\mathrm{2}} \right)......\left({i}\right) \\ $$$$\frac{{dy}}{{dX}}={x}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{2}{X}\right) \\ $$$$\frac{{dy}}{{dX}}={x}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{2}{X}\right)={x}−\frac{{gx}^{\mathrm{2}} }{{u}^{\mathrm{2}} }{X} \\ $$$$\frac{{dy}}{{dX}}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}={x}−\frac{{gx}^{\mathrm{2}} }{{v}^{\mathrm{2}} }{X}\:\:\:\:\:\:\:\:\:\:\:{X}=\frac{{v}^{\mathrm{2}} }{{gx}}\:\:\:\:\:\:.....\left({ii}\right) \\ $$$${y}={xX}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+{X}^{\mathrm{2}} \right) \\ $$$${y}_{{max}} ={x}\frac{{v}^{\mathrm{2}} }{{gx}}\:−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}}\:\right)=\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} } \\ $$$${y}_{{max}} =\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} } \\ $$

Commented by Spillover last updated on 08/Jul/24

method 1

$${method}\:\mathrm{1} \\ $$

Commented by peter frank last updated on 08/Jul/24

thanks

$$\mathrm{thanks} \\ $$

Answered by Spillover last updated on 08/Jul/24

from equation of trajactory y=xtan θ−((gx^2 )/(2v^2 ))sec^2 θ ....(i) (dy/dθ)=xsec^2 θ−((gx^2 )/(2v^2 ))2sec^2 θtan θ 0=xsec^2 θ−((gx^2 )/(2v^2 ))2sec^2 θtan θ tan θ=(v^2 /(gx^2 )) .....(ii) from (i) y=xtan θ−((gx^2 )/(2v^2 ))sec^2 θ y_(max) =x(v^2 /(gx^2 )) −((gx^2 )/(2v^2 ))(1+(v^2 /(gx^2 )) ) y_(max) =x.(v^2 /(gx^2 ))−((gx^2 )/(2v^2 ))(1+(v^2 /(gx^2 ))) y_(max) =x.(v^2 /(gx^2 ))−((gx^2 )/(2v^2 ))(1+(v^2 /(gx^2 ))) y_(max) =x.(v^2 /(gx^2 ))−((gx^2 )/(2v^2 ))−((gx^2 v^2 )/(2v^2 gx^2 )) y_(max) =((2v^4 −g^2 x^2 −v^4 )/(2v^2 g)) y_(max) =((v^4 −g^2 x^2 )/(2v^2 g))

$${from}\:{equation}\:{of}\:{trajactory} \\ $$$${y}={x}\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{sec}\:^{\mathrm{2}} \theta\:\:\:\:....\left({i}\right) \\ $$$$\frac{{dy}}{{d}\theta}={x}\mathrm{sec}\:^{\mathrm{2}} \theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta \\ $$$$\mathrm{0}={x}\mathrm{sec}\:^{\mathrm{2}} \theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta \\ $$$$\mathrm{tan}\:\theta=\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\:\:\:\:.....\left({ii}\right) \\ $$$${from}\:\left({i}\right) \\ $$$${y}={x}\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{sec}\:^{\mathrm{2}} \theta\:\:\:\: \\ $$$${y}_{{max}} ={x}\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\:\:−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\:\right) \\ $$$${y}_{{max}} ={x}.\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\right) \\ $$$${y}_{{max}} ={x}.\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\right) \\ $$$${y}_{{max}} ={x}.\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} {gx}^{\mathrm{2}} } \\ $$$${y}_{{max}} =\frac{\mathrm{2}{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} −{v}^{\mathrm{4}} }{\mathrm{2}{v}^{\mathrm{2}} {g}} \\ $$$${y}_{{max}} =\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} {g}} \\ $$$$ \\ $$

Commented by Spillover last updated on 08/Jul/24

method 2

$${method}\:\mathrm{2} \\ $$

Answered by mr W last updated on 08/Jul/24

Commented by mr W last updated on 08/Jul/24

x=v cos θ t ⇒t=(x/(v cos θ)) h=v sin θ t−((gt^2 )/2)=x tan θ−((gx^2 (1+tan^2 θ))/(2v^2 )) let λ=tan θ h=xλ−((gx^2 (1+λ^2 ))/(2v^2 )) (dh/dλ)=x−((gx^2 2λ)/(2v^2 ))=0 ⇒λ=(v^2 /(gx)), i.e. θ=tan^(−1) (v^2 /(gx)) h_(max) =x×(v^2 /(gx))−((gx^2 )/(2v^2 ))(1+(v^4 /(g^2 x^2 ))) ⇒h_(max) =((v^4 −g^2 x^2 )/(2gv^2 )) ✓ with x≤ maximum range (v^2 /g)

$${x}={v}\:\mathrm{cos}\:\theta\:{t}\:\Rightarrow{t}=\frac{{x}}{{v}\:\mathrm{cos}\:\theta} \\ $$$${h}={v}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}={x}\:\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}{v}^{\mathrm{2}} } \\ $$$${let}\:\lambda=\mathrm{tan}\:\theta \\ $$$${h}={x}\lambda−\frac{{gx}^{\mathrm{2}} \left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{\mathrm{2}{v}^{\mathrm{2}} } \\ $$$$\frac{{dh}}{{d}\lambda}={x}−\frac{{gx}^{\mathrm{2}} \mathrm{2}\lambda}{\mathrm{2}{v}^{\mathrm{2}} }=\mathrm{0}\: \\ $$$$\Rightarrow\lambda=\frac{{v}^{\mathrm{2}} }{{gx}},\:{i}.{e}.\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{v}^{\mathrm{2}} }{{gx}} \\ $$$${h}_{{max}} ={x}×\frac{{v}^{\mathrm{2}} }{{gx}}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{4}} }{{g}^{\mathrm{2}} {x}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{h}_{{max}} =\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} }\:\checkmark \\ $$$${with}\:{x}\leqslant\:{maximum}\:{range}\:\frac{{v}^{\mathrm{2}} }{{g}} \\ $$

Commented by peter frank last updated on 08/Jul/24

appriciate sir

$$\mathrm{appriciate}\:\mathrm{sir} \\ $$

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