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Question Number 209098 by Spillover last updated on 01/Jul/24

Answered by A5T last updated on 02/Jul/24

$$4=\frac{v_0^2{sin}^2\theta }{2g}\Rightarrow sin\theta =\frac{\sqrt{8g}}{v_0}$$$$R=v_{0}cos\theta \times 2t=v_0\sqrt{1-\frac{8g}{v_0^2}}\times 2t=2t\sqrt{v_0^2-8g}$$$$Rismaximumwhentismaximum$$$$4=\frac{{gt}^2}{2}\Rightarrow t=\sqrt{\frac{8}{g}}$$$$\Rightarrow R=2\sqrt{8}\times \sqrt{\frac{v_0^2}{g}-8}=4\sqrt{2}\left(\sqrt{\frac{v_0^2}{g}-8}\right)$$

Commented by Spillover last updated on 03/Jul/24

$$$$that's right, the question will be wrong after writing 8, they wrote g

Answered by Spillover last updated on 03/Jul/24

$$H_{max}=\frac{v_y^2}{2g}{v}_y=v\sin \theta H_{max}=\frac{{\left(v\sin \theta \right)}^2}{2g}$$$$H_{max}⩽4metres$$$$\frac{{\left(v\sin \theta \right)}^2}{2g}⩽4\sin {}^2\theta ⩽\frac{8g}{v^2}$$$$\sin {}^2\theta +\cos {}^2\theta =1$$$$\cos \theta =\sqrt{1-\frac{8g}{v^2}}$$$$FromRange$$$$R=\frac{v^2\sin 2\theta }{g}R=\frac{v^{2}2\sin \theta \cos \theta }{g}$$$$R=\frac{v^{2}2\sqrt{\left(\frac{8g}{v^2}\right)}.\left(\sqrt{1-\frac{8g}{v^2}}\right)}{g}$$$$R=\frac{v^{2}2.\left(\sqrt{\frac{8g}{v^2}(1-\frac{8g}{v^2}}\right)}{g}$$$$R=\frac{v^{2}2.\left(\sqrt{\frac{8g}{v^2}(\frac{v^2-8g}{v^2}}\right)}{g}$$$$R=\frac{2\sqrt{8g}\sqrt{v^2-8g}}{g}$$$$.......$$A particle is projected with initial speed V at an angle θ to the horizontal. The maximum height H of the projectile must be less than or equal to the height of the tunnel, which is 4.0 meters.

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