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Question Number 94922 by peter frank last updated on 21/May/20

Answered by mr W last updated on 22/May/20

$$timet:$$$$-h=V_0\sin \theta t-\frac{1}{2}{gt}^2$$$$\Rightarrow t=\frac{V_0}{g}(\sin \theta +\sqrt{{\sin }^2\theta +\frac{2gh}{V_0^2}})$$$$rangeR:$$$$R=V_0\cos \theta t=\frac{V_0^2}{g}\cos \theta (\sin \theta +\sqrt{{\sin }^2\theta +\frac{2gh}{V_0^2}})$$$$f\left(\theta \right)=\cos \theta (\sin \theta +\sqrt{{\sin }^2\theta +\lambda })$$$$with\lambda =\frac{2gh}{V_0^2}$$$$suchthatRismaximum,$$$$\frac{df\left(\theta \right)}{d\theta }=0$$$$\cos \theta (\cos \theta +\frac{\sin \theta \cos \theta }{\sqrt{{\sin }^2\theta +\lambda }})-\sin \theta (\sin \theta +\sqrt{{\sin }^2\theta +\lambda })=0$$$$\cos 2\theta +\frac{\sin \theta {\cos }^2\theta }{\sqrt{{\sin }^2\theta +\lambda }}=\sin \theta \sqrt{{\sin }^2\theta +\lambda }$$$$\cos 2\theta \sqrt{{\sin }^2\theta +\lambda }=(\lambda -\cos 2\theta )\sin \theta$$$${\cos }^{2}2\theta ({\sin }^2\theta +\lambda )=({\lambda }^2-2\lambda \cos 2\theta +{\cos }^{2}2\theta ){\sin }^2\theta$$$$2\lambda {\cos }^{2}2\theta =({\lambda }^2-2\lambda \cos 2\theta )(1-\cos 2\theta )$$$$(2+\lambda )\cos 2\theta =\lambda$$$$\cos 2\theta =\frac{\lambda }{\lambda +2}=\frac{1}{1+\frac{V_0^2}{gh}}=\frac{gh}{V_0^2+gh}$$$$\Rightarrow \mathit{\theta }=\frac{1}{2}{\mathbf{cos}}^{-1}\frac{\mathbf{gh}}{{\mathbf{V}}_0^2+\mathbf{gh}}$$

Commented by peter frank last updated on 24/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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