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Heights and Distances

Example:PSLV is launched from Shri Harikota . A observer standing at a distance of 60 $km$ from the point of launch of PSVL observed that the PSLV makes an angle of elevation of 30° after 15 minutes from the launch and 60° after 30 minutes. Find the distance travelled by the PSLV in the 2nd 15 minutes ans the speed of the PSLV ?

Let the PSLV travelled a distance of $x_1$ in 15 minutes.
In ∆PAB, $tan 30°$ = ${AB}/{PA}$ = ${x_1}/{60}$ 
         $1/√3$	 = ${x_1}/{60}$		   
         ${x_1} = 60/√3$=  $20√3$ $km$
		 
Let the PSLV travelled a distance of $x_2$ in 30 minutes.
Therefore , in ∆PAC , $tan 60°$ = ${AC}/{PA}$ = $x_2/60$ 
						$√3$ = $x_2/60$
						  $x_2$ = $60√3$ $km$

Distance travelled by PSLV in 2nd 15 minutes = $x_2 - x_1$
                                              = $60√3 - 20√3 $
											  = $40√3$ $km$
Average speed of PSLV after 30 minutes = ${60√3}/{1/2}$ $km$/${hr}$	
                                       = $120√3$ $km$/${hr}$

Chapters For Class X- CBSE