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Question Number 186345 by ajfour last updated on 03/Feb/23

Answered by mr W last updated on 04/Feb/23

Commented by mr W last updated on 04/Feb/23

$$\sqrt{2gb}=e\sqrt{{\left(u\sin \theta \right)}^2+2ga}$$$$\Rightarrow u\sin \theta =\sqrt{2gb(\frac{1}{e^2}-\frac{a}{b})}$$$$t=\frac{c}{u\cos \theta }=\frac{c\tan \theta }{u\sin \theta }=\frac{c\tan \theta }{\sqrt{2gb(\frac{1}{e^2}-\frac{a}{b})}}$$$$b=\frac{{gt}_2^2}{2}\Rightarrow t_2=\sqrt{\frac{2b}{g}}$$$$t_1=t-t_2=\frac{c\tan \theta }{\sqrt{2gb(\frac{1}{e^2}-\frac{a}{b})}}-\sqrt{\frac{2b}{g}}$$$$\Rightarrow t_1=(\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1)\sqrt{\frac{2b}{g}}$$$$a=u\sin \theta t_1+\frac{{gt}_1^2}{2}$$$$a=[\sqrt{2gb(\frac{1}{e^2}-\frac{a}{b})}+\frac{g}{2}(\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1)\sqrt{\frac{2b}{g}}](\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1)\sqrt{\frac{2b}{g}}$$$$\frac{a}{b}=(2\sqrt{\frac{1}{e^2}-\frac{a}{b}}+\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1)(\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1)$$$$let\mu =\frac{a}{b},\lambda =2\sqrt{\frac{1}{e^2}-\frac{a}{b}},\xi =\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1$$$$\mu =(\lambda +\xi )\xi$$$${\xi }^2+\lambda \xi -\mu =0$$$$\Rightarrow \xi =\frac{\sqrt{{\lambda }^2+4\mu }-\lambda }{2}$$$$\frac{c\tan \theta }{2b\sqrt{\frac{1}{e^2}-\frac{a}{b}}}-1=\frac{1}{e}-\sqrt{\frac{1}{e^2}-\frac{a}{b}}$$$$\Rightarrow \tan \theta =\frac{2b}{c}(1+\frac{1}{e}-\sqrt{\frac{1}{e^2}-\frac{a}{b}})\sqrt{\frac{1}{e^2}-\frac{a}{b}}$$$$\Rightarrow u=\frac{\sqrt{2gb}}{\sin \theta }\sqrt{\frac{1}{e^2}-\frac{a}{b}}$$$$x=c-u\cos \theta t_2$$$$x=c-\frac{2b}{\tan \theta }\sqrt{\frac{1}{e^2}-\frac{a}{b}}$$$$\Rightarrow x=c(1-\frac{1}{1+\frac{1}{e}-\sqrt{\frac{1}{e^2}-\frac{a}{b}}})$$

Commented by mr W last updated on 04/Feb/23

Commented by mr W last updated on 04/Feb/23

Commented by ajfour last updated on 04/Feb/23

$$Thankyousir,thatssuperb$$$$agreement.Ishallattemttoo.$$

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