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Question-181522




Question Number 181522 by mr W last updated on 26/Nov/22
Commented by mr W last updated on 26/Nov/22
two objects are fired at the same time  from point A and point B as shown.  what is the smallest distance   between them after which time?
$${two}\:{objects}\:{are}\:{fired}\:{at}\:{the}\:{same}\:{time} \\ $$$${from}\:{point}\:{A}\:{and}\:{point}\:{B}\:{as}\:{shown}. \\ $$$${what}\:{is}\:{the}\:{smallest}\:{distance}\: \\ $$$${between}\:{them}\:{after}\:{which}\:{time}? \\ $$
Answered by mahdipoor last updated on 26/Nov/22
V_(AB) =V_(AO) −V_(BO) =  (40cos60,−gt+40sin60)−(−30cos45,−gt+30cos45)=  (20+15(√2),20(√3)−15(√2))  with this speed we can assumed B is fixed   and A is animated that speed is V=V_(AB)   trajectory of Ais line , when AB⊥this line ⇒ AB is min  tanθ=((20(√3)−15(√2))/(20+15(√2))) , 100×sinθ=min(AB)  ⇒θ≈18.04 ⇒ minAB≈30.97 km
$${V}_{{AB}} ={V}_{{AO}} −{V}_{{BO}} = \\ $$$$\left(\mathrm{40}{cos}\mathrm{60},−{gt}+\mathrm{40}{sin}\mathrm{60}\right)−\left(−\mathrm{30}{cos}\mathrm{45},−{gt}+\mathrm{30}{cos}\mathrm{45}\right)= \\ $$$$\left(\mathrm{20}+\mathrm{15}\sqrt{\mathrm{2}},\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{15}\sqrt{\mathrm{2}}\right) \\ $$$${with}\:{this}\:{speed}\:{we}\:{can}\:{assumed}\:{B}\:{is}\:{fixed}\: \\ $$$${and}\:{A}\:{is}\:{animated}\:{that}\:{speed}\:{is}\:{V}={V}_{{AB}} \\ $$$${trajectory}\:{of}\:{Ais}\:{line}\:,\:{when}\:{AB}\bot{this}\:{line}\:\Rightarrow\:{AB}\:{is}\:{min} \\ $$$${tan}\theta=\frac{\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{20}+\mathrm{15}\sqrt{\mathrm{2}}}\:,\:\mathrm{100}×{sin}\theta={min}\left({AB}\right) \\ $$$$\Rightarrow\theta\approx\mathrm{18}.\mathrm{04}\:\Rightarrow\:{minAB}\approx\mathrm{30}.\mathrm{97}\:{km} \\ $$
Commented by mr W last updated on 26/Nov/22
great!
$${great}! \\ $$
Answered by mr W last updated on 26/Nov/22
Commented by mr W last updated on 26/Nov/22
using relative motion:  v_x =40 cos 60°+30 cos 45°=20+15(√2)  v_y =40 sin 60°−30 sin 45°=20(√3)−15(√2)  v=(√((20+15(√2))^2 +(20(√3)−15(√2))^2 ))=10(√(25+6(√2)−6(√6)))  tan θ=((20(√3)−15(√2))/(20+15(√2)))=((4(√3)−3(√2))/(4+3(√2)))  BP=100 sin θ=((50(4(√3)−3(√2)))/( (√(25+6(√2)−6(√6)))))≈30.979 m  AP=100 cos θ=((50(4+3(√2)))/( (√(25+6(√2)−6(√6)))))  t=((50(4+3(√2)))/( 10(√((25+6(√2)−6(√6))(25+6(√2)−6(√6))))))   =((5(4+3(√2)))/( 25+6(√2)−6(√6)))≈2.194 s
$${using}\:{relative}\:{motion}: \\ $$$${v}_{{x}} =\mathrm{40}\:\mathrm{cos}\:\mathrm{60}°+\mathrm{30}\:\mathrm{cos}\:\mathrm{45}°=\mathrm{20}+\mathrm{15}\sqrt{\mathrm{2}} \\ $$$${v}_{{y}} =\mathrm{40}\:\mathrm{sin}\:\mathrm{60}°−\mathrm{30}\:\mathrm{sin}\:\mathrm{45}°=\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{15}\sqrt{\mathrm{2}} \\ $$$${v}=\sqrt{\left(\mathrm{20}+\mathrm{15}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{15}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{10}\sqrt{\mathrm{25}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{6}\sqrt{\mathrm{6}}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{20}+\mathrm{15}\sqrt{\mathrm{2}}}=\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}} \\ $$$${BP}=\mathrm{100}\:\mathrm{sin}\:\theta=\frac{\mathrm{50}\left(\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{25}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{6}\sqrt{\mathrm{6}}}}\approx\mathrm{30}.\mathrm{979}\:{m} \\ $$$${AP}=\mathrm{100}\:\mathrm{cos}\:\theta=\frac{\mathrm{50}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{25}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{6}\sqrt{\mathrm{6}}}} \\ $$$${t}=\frac{\mathrm{50}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)}{\:\mathrm{10}\sqrt{\left(\mathrm{25}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{6}\sqrt{\mathrm{6}}\right)\left(\mathrm{25}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{6}\sqrt{\mathrm{6}}\right)}} \\ $$$$\:=\frac{\mathrm{5}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)}{\:\mathrm{25}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{6}\sqrt{\mathrm{6}}}\approx\mathrm{2}.\mathrm{194}\:{s} \\ $$
Commented by mr W last updated on 26/Nov/22

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