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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

$$‘‘withaverticalvelocity{v}^{″}shouldbe$$$$‘‘withavelocity{v}^{″}.$$$$answer\frac{v^2-g^2{x}^2}{2{gv}^2}shouldbe\frac{v^4-g^2{x}^2}{2{gv}^2}.$$

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

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

Answered by Spillover last updated on 08/Jul/24

$$y=x\tan \theta -\frac{{gx}^2}{2u^2}\sec {}^2\theta$$$$u=v$$$$let\tan \theta =X$$$$y=xX-\frac{{gx}^2}{2v^2}(1+X^2)......\left(i\right)$$$$\frac{dy}{dX}=x-\frac{{gx}^2}{2v^2}\left(2X\right)$$$$\frac{dy}{dX}=x-\frac{{gx}^2}{2v^2}\left(2X\right)=x-\frac{{gx}^2}{u^2}X$$$$\frac{dy}{dX}=00=x-\frac{{gx}^2}{v^2}XX=\frac{v^2}{gx}.....\left(ii\right)$$$$y=xX-\frac{{gx}^2}{2v^2}(1+X^2)$$$$y_{max}=x\frac{v^2}{gx}-\frac{{gx}^2}{2v^2}(1+\frac{v^2}{gx})=\frac{v^4-g^2{x}^2}{2{gv}^2}$$$$y_{max}=\frac{v^4-g^2{x}^2}{2{gv}^2}$$

Commented by Spillover last updated on 08/Jul/24

$$method1$$

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

$$\mathrm{thanks}$$

Answered by Spillover last updated on 08/Jul/24

$$fromequationoftrajactory$$$$y=x\tan \theta -\frac{{gx}^2}{2v^2}\sec {}^2\theta ....\left(i\right)$$$$\frac{dy}{d\theta }=x\sec {}^2\theta -\frac{{gx}^2}{2v^2}2\sec {}^2\theta \tan \theta$$$$0=x\sec {}^2\theta -\frac{{gx}^2}{2v^2}2\sec {}^2\theta \tan \theta$$$$\tan \theta =\frac{v^2}{{gx}^2}.....\left(ii\right)$$$$from\left(i\right)$$$$y=x\tan \theta -\frac{{gx}^2}{2v^2}\sec {}^2\theta$$$$y_{max}=x\frac{v^2}{{gx}^2}-\frac{{gx}^2}{2v^2}(1+\frac{v^2}{{gx}^2})$$$$y_{max}=x.\frac{v^2}{{gx}^2}-\frac{{gx}^2}{2v^2}(1+\frac{v^2}{{gx}^2})$$$$y_{max}=x.\frac{v^2}{{gx}^2}-\frac{{gx}^2}{2v^2}(1+\frac{v^2}{{gx}^2})$$$$y_{max}=x.\frac{v^2}{{gx}^2}-\frac{{gx}^2}{2v^2}-\frac{{gx}^2{v}^2}{2v^2{gx}^2}$$$$y_{max}=\frac{2v^4-g^2{x}^2-v^4}{2v^{2}g}$$$$y_{max}=\frac{v^4-g^2{x}^2}{2v^{2}g}$$$$$$

Commented by Spillover last updated on 08/Jul/24

$$method2$$

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

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

$$x=v\cos \theta t\Rightarrow t=\frac{x}{v\cos \theta }$$$$h=v\sin \theta t-\frac{{gt}^2}{2}=x\tan \theta -\frac{{gx}^2(1+{\tan }^2\theta )}{2v^2}$$$$let\lambda =\tan \theta$$$$h=x\lambda -\frac{{gx}^2(1+{\lambda }^2)}{2v^2}$$$$\frac{dh}{d\lambda }=x-\frac{{gx}^{2}2\lambda }{2v^2}=0$$$$\Rightarrow \lambda =\frac{v^2}{gx},i.e.\theta ={\tan }^{-1}\frac{v^2}{gx}$$$$h_{max}=x\times \frac{v^2}{gx}-\frac{{gx}^2}{2v^2}(1+\frac{v^4}{g^2{x}^2})$$$$\Rightarrow h_{max}=\frac{v^4-g^2{x}^2}{2{gv}^2}✓$$$$withx⩽maximumrange\frac{v^2}{g}$$

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

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

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