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Question Number 169969 by Beginner last updated on 13/May/22

Answered by mr W last updated on 13/May/22

Commented by ajfour last updated on 13/May/22

$$u=\sqrt{2gh}$$$$\left(u\sin \theta \right)t+\frac{\left(g\sin \theta \right)t^2}{2}=s$$$$\left(u\cos \theta \right)t-\frac{\left(g\cos \theta \right)t^2}{2}=0$$$$\Rightarrow t=\frac{2u}{g}$$$$s=\frac{2u^2\sin \theta }{g}+\frac{2u^2\sin \theta }{g}$$$$s=\frac{4\left(2gh\right)\sin \theta }{g}$$$$s=8h\sin \theta$$

Commented by mr W last updated on 13/May/22

$$greatsir!$$

Commented by learner22 last updated on 13/May/22

$$\mathit{b}\mathit{r}\mathit{a}\mathit{v}\mathit{o}$$

Commented by mr W last updated on 13/May/22

Commented by mr W last updated on 13/May/22

$$assumethattheballhitstheplane$$$$again,then$$$$with\lambda =\frac{a}{h}=\frac{l\sin \theta }{h}and\xi =\frac{l}{h}wehave$$$$\cos 2\theta +\sqrt{{\cos }^{2}2\theta +\xi \sin \theta }=\frac{\xi \sin \theta }{2(1-\cos 2\theta )}$$$$\sqrt{{\cos }^{2}2\theta +\xi \sin \theta }=\frac{\xi \sin \theta }{2(1-\cos 2\theta )}-\cos 2\theta$$$$\Rightarrow \xi =8\sin \theta$$

Commented by mr W last updated on 13/May/22

$$sayimpactpointBisaabovethe$$$$ground.$$$$speedatpointB:$$$$v=\sqrt{2gh}$$$$uponpointB:$$$$x=v\sin 2\theta t$$$$y=a+v\cos 2\theta t-\frac{{gt}^2}{2}$$$$y=a+v\cos 2\theta \times \frac{x}{v\sin 2\theta }-\frac{g}{2}\times {\left(\frac{x}{v\sin 2\theta }\right)}^2$$$$y=a+\frac{x}{\tan 2\theta }-\frac{{gx}^2}{2v^2{\sin }^{2}2\theta }$$$$\Rightarrow y=a+\frac{x}{\tan 2\theta }-\frac{x^2}{4h{\sin }^{2}2\theta }$$$$aty=0:$$$$a+\frac{x}{\tan 2\theta }-\frac{x^2}{4h{\sin }^{2}2\theta }=0$$$$x^2-\frac{4h{\sin }^{2}2\theta x}{\tan 2\theta }-4ah{\sin }^{2}2\theta =0$$$$x=\frac{2h{\sin }^{2}2\theta }{\tan 2\theta }+2\sin 2\theta \sqrt{h(h{\cos }^{2}2\theta +a)}$$$$suchthattheballhitstheplaneagain,$$$$x=\frac{2h{\sin }^{2}2\theta }{\tan 2\theta }+2\sin 2\theta \sqrt{h(h{\cos }^{2}2\theta +a)}⩽b=\frac{a}{\tan \theta }$$$$\frac{2{\sin }^{2}2\theta }{\tan 2\theta }+2\sin 2\theta \sqrt{{\cos }^{2}2\theta +\frac{a}{h}}⩽\frac{a}{h\tan \theta }$$$$let\lambda =\frac{a}{h}$$$$\Rightarrow \cos 2\theta +\sqrt{{\cos }^{2}2\theta +\lambda }⩽\frac{\lambda }{2(1-\cos 2\theta )}$$$$example:\theta =30°$$$$\lambda =2.2873$$$$i.e.when\frac{a}{h}⩾2.2873,theballwill$$$$hittheplaneagainaftertheimpact.$$

Commented by Tawa11 last updated on 14/May/22

$$\mathrm{Great}\mathrm{sir}$$

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