Question

The angle of elevation of jet fighter from $A$ point on the ground is ${60}^0$. After a fight of $15$ seconds angle of elevation changes to ${30}^0$. If the jet is flying at a speed of $720$ km/hr. Find the constant height at which jet is flying. (Take $\sqrt{3}=1.732$).

• A. $2.598km$
• B. $2.598m$
• C. $280m$
• D. $2.97m$

Answer 1

The correct option is A $2.598km$

speed=$720km/h$

$=720\times \frac{5}{18}=200m/s$

distance travelled in $16sec$

$200\times 15$

now

$\tan {30}^{\circ }=\frac{h}{x+3000}=\frac{1}{\sqrt{3}}$

$\tan {60}^{\circ }=\frac{h}{x}=\sqrt{3}$

$=\frac{\frac{h}{x}}{\frac{h}{x+3000}}=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=\frac{x+3000}{x}=3=x+3000=3x=2x=3000=x=1500$

The constant height$=1500\sqrt{3}$

$=1500\times 1.732$

$=2598m=2.598km$