Question

The angle of elevation of a Jet fighter from a point on the ground is ${60}^o$. After $15$seconds flight, the angle of elevation changes to ${30}^o$. If the Jet is flying at height of $1500\sqrt{3}m$, find the speed of the Jet.

• A. $682km/hr$
• B. $861.56km/hr$
• C. $720km/hr$
• D. $934.3km/hr$

Answer 1

The correct option is C $720km/hr$

Let the speed of the plane be along x axis.
Now
$\tan {60}^0=\frac{1500\sqrt{3}}{x}=\sqrt{3}$ ...(x=horizontal distance)
Hence
$x=1500m$
Now after 15secs
$\tan {30}^0=\frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{x+△x}$
Hence
$x+△x=4500m$
$1500m+△x=4500m$
$△x=3000m$
Speed
$=\frac{△x}{△t}=\frac{3000}{15}=200m/s$
$=720km/hr$