A-projectile-projected-from-the-ground-has-its-direction-of-motion-making-an-angle-pi-4-with-the-horizontal-at-a-height-40-m-Its-initial-velocity-of-projection-is-50-m-s-the-angle-of-projection-is-

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Question Number 15591 by Tinkutara last updated on 12/Jun/17

$$\mathrm{A}\mathrm{projectile}\mathrm{projected}\mathrm{from}\mathrm{the}\mathrm{ground}$$$$\mathrm{has}\mathrm{its}\mathrm{direction}\mathrm{of}\mathrm{motion}\mathrm{making}\mathrm{an}$$$$\mathrm{angle}\frac{\pi }{4}\mathrm{with}\mathrm{the}\mathrm{horizontal}\mathrm{at}\mathrm{a}\mathrm{height}$$$$40\mathrm{m}.\mathrm{Its}\mathrm{initial}\mathrm{velocity}\mathrm{of}\mathrm{projection}\mathrm{is}$$$$50\mathrm{m}/\mathrm{s},\mathrm{the}\mathrm{angle}\mathrm{of}\mathrm{projection}\mathrm{is}?$$

Answered by mrW1 last updated on 12/Jun/17

$$\mathrm{u}=50\mathrm{m}/\mathrm{s}$$$${\mathrm{u}}_{\mathrm{x}}=50\cos \alpha$$$${\mathrm{u}}_{\mathrm{y}}=50\sin \alpha$$$$\mathrm{y}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$$$$\mathrm{x}={\mathrm{u}}_{\mathrm{x}}\mathrm{t}$$$$\tan \theta =\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}}\times \frac{1}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{{\mathrm{u}}_{\mathrm{y}}-\mathrm{gt}}{{\mathrm{u}}_{\mathrm{x}}}$$$$\mathrm{at}\mathrm{y}=40\mathrm{m}:\vartheta =\frac{\pi }{4}$$$$\tan \frac{\pi }{4}=\frac{{\mathrm{u}}_{\mathrm{y}}-\mathrm{gt}}{{\mathrm{u}}_{\mathrm{x}}}=1$$$$\mathrm{t}=\frac{{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}}{\mathrm{g}}$$$$\mathrm{y}={\mathrm{u}}_{\mathrm{y}}\frac{{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}}{\mathrm{g}}-\frac{\mathrm{g}}{2}{\left(\frac{{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}}{\mathrm{g}}\right)}^2=40$$$$\frac{{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}}{\mathrm{g}}[{\mathrm{u}}_{\mathrm{y}}-\frac{{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}}{2}]=40$$$$\frac{{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}}{\mathrm{g}}\left[\frac{{\mathrm{u}}_{\mathrm{y}}+{\mathrm{u}}_{\mathrm{x}}}{2}\right]=40$$$${\mathrm{u}}_{\mathrm{y}}^2-{\mathrm{u}}_{\mathrm{x}}^2=2\times 40\mathrm{g}$$$${\mathrm{u}}^2({\sin }^2\alpha -{\cos }^2\alpha )=80\mathrm{g}$$$$-{\mathrm{u}}^2\cos 2\alpha =80\mathrm{g}$$$$\cos 2\alpha =-\frac{80\mathrm{g}}{{\mathrm{u}}^2}=-\frac{800}{{50}^2}=-0.32$$$$\Rightarrow 2\alpha =108.7°$$$$\Rightarrow \alpha =54.3°$$

Commented by Tinkutara last updated on 12/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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