Exam

JEE Main

Year

2018

Date

15 April

Shift

Morning Shift

Mode

Online

Subject

Mathematics

Chapter

Heights and Distances

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Heights and Distances

An aeroplane flying at a constant speed, parallel to the horizontal...

An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3} \mathrm{~km}$ above it, is observed at an elevation of $60^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^{\circ}$, then the speed (in $\mathrm{km} / \mathrm{hr}$ ) of the aeroplane is

Step-by-step solution

For $\Delta \mathrm{OA}, \mathrm{A}, \mathrm{OA}_1=\frac{\sqrt{3}}{\tan 60^{{}^{\circ}}}=1 \mathrm{~km}$
For $\Delta \mathrm{OB}_1, \mathrm{~B}, \mathrm{OB}_1=\frac{\sqrt{3}}{\tan 30^{{}^{\circ}}}=3 \mathrm{~km}$. As, a distance of $3-1=2 \mathrm{~km}$ is covered in 5 seconds. Therefore the speed of the plane is $\frac{2 \times 3600}{5}=$1440 $\mathrm{~km}$/ $\mathrm{hr}$