The angle of elevation of a jet from a point P on the ground is 60degr

1.	The angle of elevation of a jet from a point P on the ground is 60degree after 15seconds the angle of elevation changes to 30degree if a jet is flying at 720km per second find the height at which the jet is flying
Answer» Question: The angle of elevation of a jet from a point P on the ground is 60°. After 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hr, find the height at which the jet is flying? ANSWER: $\bigstar{\bold{Height\:of\:jet=1500\sqrt{3}\:m}}$ Step-by-step explanation: $\Large{\underline{\rm{Given:}}}$ The jet is flying at a speed of 720 km/hr The angle of elevation changes from 60° to 30° after 15 s $\Large{\underline{\rm{To\:Find:}}}$ The height at which the jet is flying $\Large{\underline{\rm{Solution:}}}$ ⇝ Here we have to find the height at which the jet is flying. ⇝ LET the height at which the jet FLIES = AB = DC ⇝ Let the distance travelled during the 15 s be DA = CB ⇝ Converting 720 km/hr to m/s 720 km/hr = 200 m/s ⇝ Now we know that, Distance = Speed × Time ⇝ Therefore, BC = 200 × 15 BC = 3000 m ⇝ Now consider Δ ABP tan 30 = AB/PB tan 30 = AB/(PC + BC) tan 30 = AB/(PC + 3000) 1/√3 = AB/(PC + 3000) ⇝ Cross multiplying, AB√3 = PC + 3000 PC = AB√3 - 3000------(1) ⇝ Now consider Δ DCP tan 60 = DC/PC tan 60 = AB/PC (∵ DC = AB) √3 = AB/PC PC√3 = AB PC = AB/√3------(2) ⇝ Equating equation 1 and 2 AB√3 - 3000 = AB/√3 AB = √3 (AB√3 - 3000) AB = 3 AB - 3000√3 3 AB - AB = 3000√3 2AB = 3000 √3 AB = 1500 √3 ⇝ Hence the jet is flying at a height of 1500√3 m. $\boxed{\bold{Height\:of\:jet=1500\sqrt{3}\:m}}$ $\setlength{\unitlength}{1cm}\begin{picture}\thicklines \multiput(2.5,0)(4,0){2}{\line(0,1){4}}\multiput(2.5,0)(0,4){2}{\line(1,0){4}}\put(2,4.2){\bf D}\put(-0.5,0){\line(1,0){4}}\put(6.5,4.2){\bf A}\qbezier(-0.5,0)(-0.5,0)(2.5,4)\qbezier(-0.5,0)(-0.5,0)(6.5,4)\put(-1,-0.5){\bf P}\put(2.5,-0.5){\bf C}\put(6.5,-0.5){\bf B}\qbezier (-0.1,0.5)(0.5,1.5)(0.5,0)\put(0.7,0.12){$30^{\circ}$}\put(0.8,1.2){$60^{\circ}$}\put(4,4.5){\bf 15 s}\end{picture}$

The angle of elevation of a jet from a point P on the ground is 60degree after 15seconds the angle of elevation changes to 30degree if a jet is flying at 720km per second find the height at which the jet is flying

Answer»

Question:

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

ANSWER:

$\bigstar{\bold{Height\:of\:jet=1500\sqrt{3}\:m}}$

Step-by-step explanation:

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

The jet is flying at a speed of 720 km/hr
The angle of elevation changes from 60° to 30° after 15 s

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

The height at which the jet is flying

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

⇝ Here we have to find the height at which the jet is flying.

⇝ LET the height at which the jet FLIES = AB = DC

⇝ Let the distance travelled during the 15 s be DA = CB

⇝ Converting 720 km/hr to m/s

720 km/hr = 200 m/s

⇝ Now we know that,

Distance = Speed × Time

⇝ Therefore,

BC = 200 × 15

BC = 3000 m

⇝ Now consider Δ ABP

tan 30 = AB/PB

tan 30 = AB/(PC + BC)

tan 30 = AB/(PC + 3000)

1/√3 = AB/(PC + 3000)

⇝ Cross multiplying,

AB√3 = PC + 3000

PC = AB√3 - 3000------(1)

⇝ Now consider Δ DCP

tan 60 = DC/PC

tan 60 = AB/PC (∵ DC = AB)

√3 = AB/PC

PC√3 = AB

PC = AB/√3------(2)

⇝ Equating equation 1 and 2

AB√3 - 3000 = AB/√3

AB = √3 (AB√3 - 3000)

AB = 3 AB - 3000√3

3 AB - AB = 3000√3

2AB = 3000 √3

AB = 1500 √3

⇝ Hence the jet is flying at a height of 1500√3 m.

$\boxed{\bold{Height\:of\:jet=1500\sqrt{3}\:m}}$

$\setlength{\unitlength}{1cm}\begin{picture}\thicklines \multiput(2.5,0)(4,0){2}{\line(0,1){4}}\multiput(2.5,0)(0,4){2}{\line(1,0){4}}\put(2,4.2){\bf D}\put(-0.5,0){\line(1,0){4}}\put(6.5,4.2){\bf A}\qbezier(-0.5,0)(-0.5,0)(2.5,4)\qbezier(-0.5,0)(-0.5,0)(6.5,4)\put(-1,-0.5){\bf P}\put(2.5,-0.5){\bf C}\put(6.5,-0.5){\bf B}\qbezier (-0.1,0.5)(0.5,1.5)(0.5,0)\put(0.7,0.12){$30^{\circ}$}\put(0.8,1.2){$60^{\circ}$}\put(4,4.5){\bf 15 s}\end{picture}$

The angle of elevation of a jet from a point P on the ground is 60degree after 15seconds the angle of elevation changes to 30degree if a jet is flying at 720km per second find the height at which the jet is flying

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