Question

The angle of elevation of a jet fighter plane from a point A on the groundis 60. After a flight of 15 seconds, the angle of elevation changes to 30.If the jet is flying at a speed of 720 km/h. Find the constant height at which the jet is flying.

Answer 1

H = 1500√3 m = 1500×1.732 = 2598 m

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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