Question

The angle of elevation of a jetfighter from a point A on the ground is 60° After a flight of 15
seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720km/hr, find the
constant height at which the jet is flying.​

Answer 1

Solution :-

Refer to attachment for the figure

Let the position of the jet fighter be E

Angle of elevation from point A to the jet fighter ∠EAC = 60°

After 15 seconds let the position of the jet fighter be D

Angle of elevation from point A to C ∠DAC = 30°

Speed of the jet fighter = 720 Km/h = 720 * 5/18 = 200 m/s

[ Because 1 km/h = 5/18 m/s ]

Time taken to reach point D from point E = 15 seconds

We know that

Distance = Speed * Time

Distance covered by jet from point E to point D ECLD = 200 * 15 = 3000 m

BE draw perpendicular to AC

DE = BC = 3000 m [ Distance between altitudes ]

Consider ΔABE

tan 60° = Opp/Adj = EB/AB

⇒ √3 = EB/AB

⇒ AB√3 = EB

⇒ AB = EB/√3 -- EQ(1)

Consider ΔADC

tan 30° = Opp/Adj = DC/AC

⇒ 1/√3 = DC/AC

⇒ AC = DC √3

From figure

⇒ AB + BC = DC√3

⇒ AB + 3000 = DC√3

⇒ AB + 3000 = EB√3

[ Given jet is flying at a constant height DC = EB ]

⇒ AB = EB√3 - 3000 --- EQ(2)

From Eq(1) & EQ(2)

⇒ EB/√3 = EB√3 - 3000

⇒ EB = √3(EB√3 - 3000)

⇒ EB = 3EB - 3000√3

⇒ 3000√3 = 3EB - EB

⇒ 3000√3 = 2EB

⇒ EB = 1500√3

⇒ EB = 1500 * 1.732 = 2598

Therefore the constant height at which jet is flying is 2598 m.

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Answer 2

$\underline \mathbb{ANSWER}$

$\bold{\huge{\fbox{\color{Red}{In\:figure}}}}$

$\bold{\huge{\fbox{\color{Red}{Let,}}}}$

》A be the point of observation on the ground and B and C be the two positions of the jet.

》BL = CM = h metres

》AL = x metres

$\bold{\huge{\fbox{\color{Red}{Then}}}}$

$\implies\:$ Speed of jet = 720 km/hr

$= > 720 \times \frac{5}{18} m/sec$

$= > 200m/sec$

》Time taken to cover the distance BC = 15 sec

》DISTANCE BC = LM = 200×15m = 3000m

$\boxed {Distance = 3000m}$

$\implies\:$IN Δ ALB,

$\implies\:$cot 60° = AL / BL

$= > \frac{1}{ \sqrt{3} } = \frac{h}{x}$

$= > x = \frac{h}{ \sqrt{3} }$

$\implies\:$IN Δ AMC,

$\implies\:$cot 30° = AM / MC

$= > \sqrt{3} = ( \frac{al + lm}{mc} )$

$= > x + 3000 = h \sqrt{3}$

$= > x = h \sqrt{3} - 3000$

》Equating the value of x,

$= > \sqrt{3} h - 3000 = \frac{h}{ \sqrt{3} }$

$= > 3h - 3000 \sqrt{3} = h$

$= > 2h = 5196$

$= > h = 2598$

$\boxed {h=2598km}$

$\bold\green{\underline{Hence,}}$

The constant height at which the jet is flying is 2598m.

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NOTE:

REFER TO ATTACHMENT FOR FIGURE.

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