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Question Number 213890 by Tawa11 last updated on 20/Nov/24

Answered by mr W last updated on 21/Nov/24

$$h=u\sin \theta t-\frac{{gt}^2}{2}$$$$t=\frac{u\sin \theta }{g}(1\pm \sqrt{1-\frac{2gh}{u^2{\sin }^2\theta }})$$$$=\frac{45\sin 52°}{9.81}(1\pm \sqrt{1-\frac{2\times 9.81\times 12}{{45}^2{\sin }^{2}52°}})$$$$=6.87s/0.36s$$$$R=u\cos \theta t$$$$=45\times \cos 52°\times 6.87/0.36$$$$=190.4m/9.86m$$

Commented by mr W last updated on 21/Nov/24

Commented by Tawa11 last updated on 21/Nov/24

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by mr W last updated on 21/Nov/24

$$whatisright?$$

Answered by A5T last updated on 21/Nov/24

$$u_x=ucos\theta ;u_y=usin\theta ;u=45{ms}^{-1};\theta =52°$$$$H=maximumheight;R=range$$$$t=timefromHto12mabovelaunchingpoint;$$$$T=flighttime$$$$H-12=\frac{{gt}^2}{2}\Rightarrow \frac{g\left(\frac{u^2{sin}^2\theta }{g^2}\right)}{2}-12=\frac{{gt}^2}{2}$$$$\Rightarrow \frac{u^2{sin}^2\theta -24g}{2g}=\frac{{gt}^2}{2}\Rightarrow t=\frac{\sqrt{u^2{sin}^2\theta -24g}}{g}$$$$\Rightarrow T=\frac{\sqrt{u^2{sin}^2\theta -24g}}{g}+\frac{usin\theta }{g}$$$$R=u_{x}T=\frac{ucos\theta \sqrt{u^2{sin}^2\theta -24g}}{g}+\frac{u^{2}sin2\theta }{2g}$$$$Thensubstitutenecessaryvalues.$$

Commented by Tawa11 last updated on 21/Nov/24

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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