Question

the angle of elevation of jet fighter is 60 degree. after passing 15 seconds the angle of elevation found to be 30 degree.if speed is 720 km per hour .find the distance travelled by jet fighter

Answer 1

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Answer 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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