Question

The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane.

Answer 1

Height of the plane from the ground PM = RN = 1500√3 m.

Angle of elevation are 30° and 60°. 

From the figure 

tan 60° = \(\frac{PM}{QM}\)

√3 = \(\frac{1500\sqrt{3}}{QM}\)

QM = \(\frac{1500\sqrt{3}}{\sqrt{3}}\) = 1500 m 

Also tan 30° = \(\frac{RN}{QN}\)

\(\frac{1}{\sqrt{3}}\) = \(\frac{1500\sqrt{3}}{QM+MN}\)

QM + MN = 1500√3 × √3 

1500 + MN = 1500 × 3 

MN = 4500 – 1500 

MN = 3000 mts. 

∴ Distance travelled in 15 seconds = 3000 mts. 

∴ Speed of the jet plane = \(\frac{distance}{time}= \frac{3000}{15}\) = 200 m/s 

= 200 × \(\frac{18}{5}\) kmph 

= 720 kmph 

Speed = 200 m/sec. or 720 kmph.