 Krishna
0

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}