Question

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​

Answer 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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Answer 2

Answer:

$\bigstar{\bold{Height\:of\:the\:jet=2598\:m}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• The angle of elevation of the jet is 60°
• After 15 seconds the  angle of elevation is 30°
• Speed of jet = 720 km/hr = 200 m/s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The height of the jet

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we have to find the distance covered  by the truck during the time interval of 15 s.

→ We know that

   Distance = Speed × Time

→ Substituting the data,

  Distance = 200 × 15

  Distance = 3000 m

→ Hence the distance covered by the jet is 3000 m.

→ Here the distance covered = EC = DB = 3000 m

→ Now let the height of the jet be x m

→ Consider Δ EDA

  tan 60 = ED/AD

  √3 = x/AD

  AD√3 = x

  AD = x/√3

  AD = (x√3)/3 -----(1)

→ Consider ΔABC

  tan 30 = CB/AB

  tan 30 = x/AD + DB

  1/√3 = x/(AD + 3000)

  AD + 3000 = x√3

  AD = x√3 - 3000 ----(2)

→ From equations 1 and 2 , LHS are equal, therefore RHS must also be equal.

 (x√3)/3 = x√3 - 3000

  x√3 = 3( x√3 - 3000)

  x√3 = √3 ×√3 (x√3 - 3000)

→ Cancelling √3 on both sides

  x = √3 (x√3 - 3000)

  x = 3x - 3000√3

  2x = 3000√3

    x = 3000√3/2

    x = 1500 × 1.732

    x = 2598 m

→ Hence the height of the jet is 2598 m

$\boxed{\bold{Height\:of\:the\:jet=2598\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Sin A = opposite/Hypotenuse

→ Cos A = adjacent/Hypoteuse

→ Tan A = opposite/adjacent