In the given figure, a rocket is fired vertically upwards from it… - Vidyadip

Question

In the given figure, a rocket is fired vertically upwards from its launching pad $P$. It first rises $20 \ km$ vertically upwards and then $20 \ km$ at $60^\circ $ to the vertical. $PQ$ represents the first stage of the journey and $QR$ the second. $S$ is a point vertically below $R$ on the horizontal level as $P$, find$:\ a$. the height of the rocket when it is at point $R.b.$ the horizontal distance of point $S$ from $P.$

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Answer

Draw $QM \perp RS.$
Clearly, $\angle RQM = 30^\circ $

$a$. In right $\triangle RMQ$,
$\sin 30^{\circ}=\frac{R M}{R Q}$
$ \Rightarrow \frac{1}{2}=\frac{R M}{20}$
$ \Rightarrow R M=10 \ km$
$\therefore$ The height of the rocket when it is at point $R$
$= RS$
$ = RM + MS$
$ =10 \ km +20 \ km$
$ =30 \ km .$
$b$. In right $\triangle R M Q$,
$\cos 30^{\circ}=\frac{ QM }{ RQ }$
$ \Rightarrow \frac{\sqrt{3}}{2}=\frac{ QM }{20}$
$ \Rightarrow QM =10 \sqrt{3} \ km$
$\therefore$ The horizontal distance of point $S$ from $P$
$=\text { PS }$
$= QM$
$ =10 \sqrt{3} \ km .$

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