Math, asked by TOMalter99999, 6 months ago

The angle of elevation of a jet fighter from a point a A on the ground is 60°. After a flight of 15 second the angle of elevation changes to 30° . if the jet is flying at a seed of 720m\hr find the constant height​

Answers

Answered by Anonymous
2

 \textsf {\large{\underline { \underline {SOLUTION:-}}}}

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

 \bf➠Then, \angle XOA=60⁰ \: and   \: \angle \: XOB=30⁰ \\  \\

  \bf ➠Draw \:   AL \perp \: OX \: and  \: BM \perp \: OX \\  \\

 \bf \:⇰ Let \: AL=BM=h \: meters. \\  \\

 \bf \: ⇰ \: Speed \: of \: Jet  = 720 \: km/hr\\

 \bf =  \huge( \small \: 720 \times  \frac{5}{18}  \huge) \small \: m / s\\

 \bf = 200 \: m/s

 \bf \:➥ Time  \: taken \: to \: cover \: the \: distance  \: AB=15  \: sec  \\  \\

 \bf➥ Distance  \: covered =(speed×time) \\  \\

 \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \bf = (200 \times 15)m = 3000m \\  \\

 \bf \therefore LM=AB=3000m \\  \\

{ \bf \: ⇰  \: Let \: OL} = x   \:  \bf m \\  \\

 \bf➥From  \: right  \: \triangle OLA, \: we \: have: \\

 \bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h}  =  \frac{1}{ \sqrt{3} }  \\  \\

 \bf \implies \: x =  \frac{h}{ \sqrt{3} }  \:  \:  \:  \:  \:  \:  \: ....(1) \\  \\

 \bf \:➥ From  \: right  \: \triangle \: OMB,we \: have: \\

  \bf\frac{OM}{BM} =cot30⁰= \sqrt{3}  \\  \\

 \bf \implies \:  \frac{x + 3000}{h}  =  \sqrt{3}  \\  \\

 \bf[ \boxed{➯OM=OL+LM=OL+AB=(x+3000)m  \: and \: BM=h \: m]}

 \bf \implies \: x + 3000 =  \sqrt{3} h \\

 \bf \implies \: x = ( \sqrt{3} h - 3000) \:  \:  \:  \:  \:  \: ....(2) \\

⇰ Equating the value of x from (1) and (2),we get;

 \bf \:  \sqrt{3} h - 3000  =  \frac{h}{ \sqrt{3} }  \\  \\

 \implies \bf \: 3h  - 3000 \sqrt{3}  = h \\

 \bf \implies \: 2h = (3000 \times  \sqrt{3} ) = (3000 \times 1.762) \\

 \bf \implies \: h = (3000 \times 0.866) = 2598  \\  \\

 \bf \:➥ Hence \: ,the \: required  \: height \: is \: \boxed{ \boxed{ \bf2598 \: m.}} \\  \\

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Answered by BARFI20000
1

 \textsf {\large{\underline { \underline {SOLUTION:-}}}}

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

 \bf➠Then, \angle XOA=60⁰ \: and   \: \angle \: XOB=30⁰ \\  \\

  \bf ➠Draw \:   AL \perp \: OX \: and  \: BM \perp \: OX \\  \\

 \bf \:⇰ Let \: AL=BM=h \: meters. \\  \\

 \bf \: ⇰ \: Speed \: of \: Jet  = 720 \: km/hr\\

 \bf =  \huge( \small \: 720 \times  \frac{5}{18}  \huge) \small \: m / s\\

 \bf = 200 \: m/s

 \bf \:➥ Time  \: taken \: to \: cover \: the \: distance  \: AB=15  \: sec  \\  \\

 \bf➥ Distance  \: covered =(speed×time) \\  \\

 \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \bf = (200 \times 15)m = 3000m \\  \\

 \bf \therefore LM=AB=3000m \\  \\

{ \bf \: ⇰  \: Let \: OL} = x   \:  \bf m \\  \\

 \bf➥From  \: right  \: \triangle OLA, \: we \: have: \\

 \bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h}  =  \frac{1}{ \sqrt{3} }  \\  \\

 \bf \implies \: x =  \frac{h}{ \sqrt{3} }  \:  \:  \:  \:  \:  \:  \: ....(1) \\  \\

 \bf \:➥ From  \: right  \: \triangle \: OMB,we \: have: \\

  \bf\frac{OM}{BM} =cot30⁰= \sqrt{3}  \\  \\

 \bf \implies \:  \frac{x + 3000}{h}  =  \sqrt{3}  \\  \\

 \bf[ \boxed{➯OM=OL+LM=OL+AB=(x+3000)m  \: and \: BM=h \: m]}

 \bf \implies \: x + 3000 =  \sqrt{3} h \\

 \bf \implies \: x = ( \sqrt{3} h - 3000) \:  \:  \:  \:  \:  \: ....(2) \\

⇰ Equating the value of x from (1) and (2),we get;

 \bf \:  \sqrt{3} h - 3000  =  \frac{h}{ \sqrt{3} }  \\  \\

 \implies \bf \: 3h  - 3000 \sqrt{3}  = h \\

 \bf \implies \: 2h = (3000 \times  \sqrt{3} ) = (3000 \times 1.762) \\

 \bf \implies \: h = (3000 \times 0.866) = 2598  \\  \\

 \bf \:➥ Hence \: ,the \: required  \: height \: is \: \boxed{ \boxed{ \bf2598 \: m.}} \\  \\

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