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Question Number 181522 by mr W last updated on 26/Nov/22

Commented by mr W last updated on 26/Nov/22

$$twoobjectsarefiredatthesametime$$$$frompointAandpointBasshown.$$$$whatisthesmallestdistance$$$$betweenthemafterwhichtime?$$

Answered by mahdipoor last updated on 26/Nov/22

$$V_{AB}=V_{AO}-V_{BO}=$$$$(40cos60,-gt+40sin60)-(-30cos45,-gt+30cos45)=$$$$(20+15\sqrt{2},20\sqrt{3}-15\sqrt{2})$$$$withthisspeedwecanassumedBisfixed$$$$andAisanimatedthatspeedisV=V_{AB}$$$$trajectoryofAisline,whenAB\mathrm{\perp }thisline\Rightarrow ABismin$$$$tan\theta =\frac{20\sqrt{3}-15\sqrt{2}}{20+15\sqrt{2}},100\times sin\theta =min\left(AB\right)$$$$\Rightarrow \theta \approx 18.04\Rightarrow minAB\approx 30.97km$$

Commented by mr W last updated on 26/Nov/22

$$great!$$

Answered by mr W last updated on 26/Nov/22

Commented by mr W last updated on 26/Nov/22

$$usingrelativemotion:$$$$v_x=40\cos 60°+30\cos 45°=20+15\sqrt{2}$$$$v_y=40\sin 60°-30\sin 45°=20\sqrt{3}-15\sqrt{2}$$$$v=\sqrt{{(20+15\sqrt{2})}^2+{(20\sqrt{3}-15\sqrt{2})}^2}=10\sqrt{25+6\sqrt{2}-6\sqrt{6}}$$$$\tan \theta =\frac{20\sqrt{3}-15\sqrt{2}}{20+15\sqrt{2}}=\frac{4\sqrt{3}-3\sqrt{2}}{4+3\sqrt{2}}$$$$BP=100\sin \theta =\frac{50(4\sqrt{3}-3\sqrt{2})}{\sqrt{25+6\sqrt{2}-6\sqrt{6}}}\approx 30.979m$$$$AP=100\cos \theta =\frac{50(4+3\sqrt{2})}{\sqrt{25+6\sqrt{2}-6\sqrt{6}}}$$$$t=\frac{50(4+3\sqrt{2})}{10\sqrt{(25+6\sqrt{2}-6\sqrt{6})(25+6\sqrt{2}-6\sqrt{6})}}$$$$=\frac{5(4+3\sqrt{2})}{25+6\sqrt{2}-6\sqrt{6}}\approx 2.194s$$

Commented by mr W last updated on 26/Nov/22

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