Question

A jet fighter plane is flying horizontally at speed of 720 km/h. from a point A on the ground, the angle of elevation of the plane is 60 degree. after flight of 15 second the angle of elevation changes to 30 degree. find the height at which the jet is flying

Answer 1

let A be the observation point
and let B and C be the two  positions of jet fighter.it takes 15 sec to reach c from b.
let the height of the jet from the ground be BE=CD=720 km/per h=720000m / hour
distance=speed x distance 
therefore BC
720000 x 15/15 x 15=3000 m = 3 km

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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