Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The angle of elevation of a jet plane from a point A on the ground is 60o. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30o. If the jet plane is flying at a constant height, then its

The angle of elevation of a jet plane from a point A on the ground is 60o. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30o. If the jet plane is flying at a constant height, then its height is :
Option: 1 1200\sqrt{3}\; m
Option: 2 3600\sqrt{3}\; m
Option: 3 1800\sqrt{3}\; m
Option: 4 2400\sqrt{3}\; m

Answers (1)

best_answer

 

\begin{aligned} &\tan 60^{\circ}=\frac{\mathrm{h}}{\mathrm{y}}\\ &\sqrt{3}=\frac{h}{y} \Rightarrow h=\sqrt{3} y\qquad\ldots(1) \end{aligned}

\begin{aligned} &\tan 30^{\circ}=\frac{\mathrm{h}}{\mathrm{x}+\mathrm{y}}\\ &\frac{1}{\sqrt{3}}=\frac{h}{x+y} \Rightarrow \sqrt{3} h=x+y\qquad\ldots(2) \end{aligned}

\\\text { Speed } 432 \mathrm{~km} / \mathrm{h} \Rightarrow \frac{432 \times 20}{60 \times 60} \Rightarrow \frac{12}{5} \mathrm{~km} \\ \\\sqrt{3} \mathrm{~h}=\frac{12}{5}+\mathrm{y} \\ \\\sqrt{3} \mathrm{~h}-\frac{12}{5}=\mathrm{y}

From (1)

\\h=\sqrt{3}\left[\sqrt{3} h-\frac{12}{5}\right] \\ \\h=3 h-\frac{12 \sqrt{3}}{5} \\ \\h=\frac{6 \sqrt{3}}{5} \mathrm{~km} \\ \\h=1200 \sqrt{3} \mathrm{~m}

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

DASA, CSAB Special round 2 seat allotment result 2025 declared on csab.nic.in

Latest Question