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Question Number 174823 by ajfour last updated on 11/Aug/22

Commented by ajfour last updated on 11/Aug/22

$$Findmaximumheighttheball$$$$canclimbupthewall.$$

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

$$t_1=\frac{b}{u\cos \theta }$$$$v_1=u\sin \theta -\frac{gb}{u\cos \theta }⩾0$$$$\sin 2\theta ⩾\frac{2gb}{u^2}$$$$\Rightarrow \frac{1}{2}{\sin }^{-1}\frac{2gb}{u^2}⩽\theta ⩽\frac{\pi }{2}-\frac{1}{2}{\sin }^{-1}\frac{2gb}{u^2}$$$$\Rightarrow u⩾\sqrt{2gb}$$$$max.heightreached:$$$$h=\frac{{\left(u\sin \theta \right)}^2}{2g}=\frac{u^2(1-\cos 2\theta )}{4g}$$$$⩽\frac{u^2}{4g}[1+\sqrt{1-{\left(\frac{2gb}{u^2}\right)}^2}]=h_{max}$$

Commented by ajfour last updated on 13/Aug/22

$$thankssir,perfect!$$

Commented by Tawa11 last updated on 14/Aug/22

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

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