A-projectile-is-fired-with-velocity-v-o-such-that-it-passes-through-two-points-both-a-distance-h-above-the-horizontal-show-that-if-the-gun-is-adjusted-for-the-maximum-range-of-the-separation-of-two

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Question Number 177548 by peter frank last updated on 07/Oct/22

$$\mathrm{A}\mathrm{projectile}\mathrm{is}\mathrm{fired}\mathrm{with}\mathrm{velocity}\left({\mathrm{v}}_{\mathrm{o}}\right)$$$$\mathrm{such}\mathrm{that}\mathrm{it}\mathrm{passes}\mathrm{through}\mathrm{two}$$$$\mathrm{points}\mathrm{both}\mathrm{a}\mathrm{distance}\left(\mathrm{h}\right)\mathrm{above}\mathrm{the}$$$$\mathrm{horizontal}.\mathrm{show}\mathrm{that}\mathrm{if}\mathrm{the}\mathrm{gun}\mathrm{is}\mathrm{adjusted}$$$$\mathrm{for}\mathrm{the}\mathrm{maximum}\mathrm{range}\mathrm{of}\mathrm{the}$$$$\mathrm{separation}\mathrm{of}\mathrm{two}\mathrm{position}\mathrm{is}$$$$\mathbf{d}=\frac{{\mathbf{v}}_{\mathbf{o}}\sqrt{{\mathbf{v}}_{\mathbf{o}}^2-4\mathbf{gh}}}{\mathbf{g}}$$

Answered by blackmamba last updated on 07/Oct/22

$$y=y_o+x\tan \theta -\frac{{gx}^2}{2v_o^2\cos {}^2\theta }$$$$since\theta =45°,y=h,y_o=0$$$$thenh=0+x\tan 45°-\frac{{gx}^2}{2v_o^2\cos {}^{2}45°}$$$$b=x-\frac{{gx}^2}{v_o^2}or{x}^2-\frac{v_o^2}{g}.x+\frac{v_o^2}{g}h=0$$$$x_1+x_2=\frac{v_o^2}{g}$$$$\Rightarrow {(x_1-x_2)}^2={(x_1+x_2)}^2-4x_1{x}_2$$$$\Rightarrow d^2={\left(\frac{v_o^2}{g}\right)}^2-4\left(\frac{v_o^{2}h}{g}\right)$$$$\Rightarrow d=\sqrt{{\left(\frac{v_o^2}{g}\right)}^2(v_o^2-4gh})$$$$\Rightarrow d=\frac{v_o\sqrt{v_o^2-4gh}}{g}$$

Commented by Tawa11 last updated on 07/Oct/22

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

Answered by Spillover last updated on 07/Oct/22

$$\mathrm{x}={\mathrm{v}}_{\mathrm{o}}\cos \theta \mathrm{t}\mathrm{y}={\mathrm{v}}_{\mathrm{o}}\sin \theta \mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$$$$\mathrm{R}=\frac{{\mathrm{v}}_{\mathrm{o}}^2\sin 2\theta }{\mathrm{g}}\theta =\frac{\pi }{4}$$$$\mathrm{x}=\frac{{\mathrm{v}}_{\mathrm{o}}\mathrm{t}}{\sqrt{2}}\mathrm{y}=\frac{{\mathrm{v}}_{\mathrm{o}}\mathrm{t}}{\sqrt{2}}-\frac{1}{2}{\mathrm{gt}}^2$$$${\mathrm{x}}^2-\frac{{\mathrm{v}}_{\mathrm{o}}^2\mathrm{x}}{\mathrm{g}}+\frac{{\mathrm{v}}_{\mathrm{o}}^2\mathrm{h}}{\mathrm{g}}=0$$$${\mathrm{x}}_1=\frac{{\mathrm{v}}_{\mathrm{o}}^2}{2\mathrm{g}}-\frac{{\mathrm{v}}_{\mathrm{o}}}{2\mathrm{g}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-4\mathrm{gh}}$$$${\mathrm{x}}_2=\frac{{\mathrm{v}}_{\mathrm{o}}^2}{2\mathrm{g}}-\frac{{\mathrm{v}}_{\mathrm{o}}}{2\mathrm{g}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-4\mathrm{gh}}$$$$\mathrm{d}={\mathrm{x}}_2-{\mathrm{x}}_1$$$$\mathrm{d}=\frac{{\mathrm{v}}_{\mathrm{o}}}{\mathrm{g}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-4\mathrm{gh}}★★★$$$$$$

Commented by peter frank last updated on 07/Oct/22

$$\mathrm{more}\mathrm{clarification}\mathrm{please}\mathrm{step}3?$$$$\mathrm{and}\mathrm{how}\theta =\frac{\pi }{4}?\mathrm{diagram}\mathrm{please}?$$$$$$

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