Question

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 1

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.

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

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

   

Answer 2

Corrrect Question:-

The angle of elevation of a jet from a point P on the ground is 60° after 15sec the angle of elevation changes to 30° if a jet is flying at 720km/hr find the height at which the jet is flying

Given:-

✬Angle of Elevation of a Jet = 60°

✬Angle of elevation of a Jet after 15sec = 30°

✬Speed of Jet = 720km/h

Find:-

✭Height ot which jet is flying

Diagram:-

Here, Let A and B initial and final position of the jet respectively and P be the observation point. The angle of elevation are 30° and 60° respectively.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(1,2)(1,2)(5,5)\qbezier(5,5)(5,5)(5,2)\put(1,2){\line(1,0){7}}\qbezier(8,5)(8,5)(8,2)\qbezier(1,2)(1,2)(8,5)\put(5,5){\line(1,0){3}}\qbezier(2,2)(3,2.6)(2,2.7)\qbezier(1.7,2)(1.8,2.7)(1.7,2) \put(2.5,2.2){\sf \bf{{60}^{ \circ}} } \put(1.8,2.1){\sf \bf{{30}^{ \circ}} } \put(5.2,3.9){\vector(0,1){1}}\put(5.2,3.5){\vector(0, - 1){1.4}}\put(8.2,3.9){\vector(0,1){1}}\put(8.2,3.5){\vector(0, - 1){1.4}}\put(4.2,1.5){\vector( - 1,0){3}}\put(6.8,1.5){\vector(1,0){1.4}} \put(5.1,3.5){\sf \bf{h m}}\put(8,3.6){\sf \bf{h m}} \put(4.3,1.5){\sf \bf{Ground Level}} \put(4.8,5.1){\sf \bf{ A }} \put(8,5.1){\sf \bf{B }}\put(4.8,1.7){\sf \bf{C}}\put(8.3,1.8){\sf \bf{D}}\put(0.7,1.8){\sf \bf{P}} \end{picture}$

Solution:-

✒Time taken to travel from A to B = 15sec

▶$\sf \bigg\lgroup{\dfrac{15}{60 \times 60}}\bigg \rgroup hrs$

▶$\sf \bigg\lgroup{\dfrac{15}{3600}}\bigg \rgroup hrs$

▶$\sf\dfrac{15}{3600}hrs$

Since, we know

$\large{\underline{\boxed{\sf Distance = Speed \times Time}}}$

where,

• Speed = 720km/hr
• Time = 15/3600 hrs

So,

$:\to\sf AB = 720 \times \dfrac{15}{3600}$

$:\to\sf AB = \dfrac{10800}{3600}$

$:\to\sf AB = 3km$

$:\to\sf AB = 3000m$

$:\implies\sf AB = CD = 3000m$

Now, In right ∆ACP

$\sf \to \dfrac{P}{B} = \tan{60}^{ \circ}$

$\sf \to \dfrac{AC}{PC} = \tan{60}^{ \circ}$

where,

• AC = h m
• tan 60° = √3

So,

$\sf \multimap \dfrac{h}{PC} = \sqrt{3}$

$\sf \multimap PC\sqrt{3} = h$

$\sf \multimap PC = \dfrac{h}{ \sqrt{3} } \qquad \huge{.....1}$

$\rule{299}{2}$

Now, In right ∆BDP

$\sf \to \dfrac{P}{B} = \tan{30}^{ \circ}$

$\sf \to \dfrac{BD}{PD} = \tan{30}^{ \circ}$

where,

• BD = h m
• PD = PC + CD
• tan 30° = 1/√3
• CD = 3000m

So,

$\sf \multimap \dfrac{h}{PC + CD} = \dfrac{1}{\sqrt{3}}$

Using eq. (1)

$\sf \multimap \dfrac{h}{ \dfrac{h}{ \sqrt{3} } + 3000} = \dfrac{1}{\sqrt{3}}$

$\sf \multimap \dfrac{h}{ \dfrac{h + 3000 \sqrt{3} }{ \sqrt{3} }} = \dfrac{1}{\sqrt{3}}$

$\sf \multimap \dfrac{h \times \sqrt{3} }{h + 3000 \sqrt{3}} = \dfrac{1}{\sqrt{3}}$

$\sf \multimap \dfrac{\sqrt{3}h}{h + 3000 \sqrt{3}} = \dfrac{1}{\sqrt{3}}$

$\sf \multimap h + 3000 \sqrt{3} \times 1 = \sqrt{3} \times \sqrt{3}h$

$\sf \multimap h + 3000 \sqrt{3}= 3h$

$\sf \multimap 3000 \sqrt{3}= 3h - h$

$\sf \multimap 3000 \sqrt{3}=2h$

$\sf \multimap \dfrac{3000 \sqrt{3}}{2}=h$

$\sf \multimap 1500 \sqrt{3}=h$

$\sf \multimap 1500 \times (1.732)=h \qquad \lgroup{using \: \sqrt{3} = 1.732} \rgroup$

$\sf \multimap 2598.07m=h$

$\sf \multimap 2598m=h$

$\sf \therefore h = 2598m$

Hence, the height at which jet is flying will be 2598m