Question

The angle of elevation of a jet fighter from a point A on the ground is ${60}^{\circ }$. After a flight of 15 seconds, the angle of elevation changes to ${30}^{\circ }$. If the jet is flying at a speed of 720 km/hour, find the constant height at which the jet is flying. [Use $\sqrt{3}=1.732$]

• A. 3598 m
• B. 2598 m
• C. 2798 m
• D. 3098 m

Answer 1

The correct option is B

2598 m

Let O be the point of observation on the ground OX.

Let A and B be the two positions of the jet.



Then, $\angle XOA={60}^{\circ }$ and $\angle XOB={30}^{\circ }$

Draw $AL\perp OX$ and $BM\perp OX$

Let AL = BM = h metres

Speed of the jet

$=(720\times \frac{5}{18})m/s$

$=200m/s$

Time taken to cover the distance AB = 15 s

Distance covered $=speed\times time$

$=200m/s\times 15s$

$=200\times 15m=3000m$

$\therefore$ LM = AB = 3000 m

Let OL = x metres

From right $\Delta OLA$, we have

$\frac{OL}{AL}=cot{60}^{\circ }=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{x}{h}=\frac{1}{\sqrt{3}}\Rightarrow x=\frac{h}{\sqrt{3}}$ ............ (i)

From right $\Delta OMB$, we have

$\frac{OM}{BM}=cot{30}^{\circ }=\sqrt{3}$

$\Rightarrow \frac{(x+3000)}{h}=\sqrt{3}\Rightarrow x=(h\sqrt{3}-3000)$ ........... (ii)

Equating the values of x from (i) and (ii), we get

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

$\Rightarrow 2h=3000\sqrt{3}\Rightarrow h=1500\sqrt{3}=1500\times 1.732=2598$

Hence the required height is 2598 m