Step 1: According to the given information construct an imaginary figure.
Let P and R be the initial and final positions of the plane respectively
Height of the jet plane PQ = RS = 1500 \sqrt{3}
The angle of elevation of a jet plane from a point A on the ground
\angle PAQ = 60\degree
The angle of elevation after 15 seconds
\angle RAS = 30\degree
Step 2: Calculate the distance traveled by the plane
Consider the right triangle APQ
\tan \theta = \frac{opposite}{adjacent} = \frac{PQ}{AQ}
\tan 60\degree = \frac{1500\sqrt{3}}{AQ}
\sqrt{3} = \frac{1500\sqrt{3}}{AQ} ,,,,,,,,,,,,,,,,,,,, ( \because \tan 60\degree = \sqrt{3})
AQ = \frac{1500\sqrt{3}}{\sqrt{3}}
AQ = 1500 m
Consider the right triangle ARS
\tan \theta = \frac{RS}{AS}
\tan 30\degree = \frac{1500\sqrt{3}}{AS}
\frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{AS} ( \because \tan 30\degree = \frac{1}{\sqrt{3}} )
AS = 1500\sqrt{3}*\sqrt{3}
AS = 1500 * 3
AS = 4500 m
From the figure
Distance traveled PR = QS = AS - AQ
= 4500 - 1500
= 3000m
Step 3:Find the speed of the jet plane
NOTE: Speed = \frac{distance}{time}
Time = 15 seconds
Distance = 3000 m
Speed = \frac{3000}{15}
Speed = 200 \frac{meter}{sec}
