The angle of elevation of a jet fighter from a point O on the ground is $60^{\circ}$. After a flight of 10 seconds, the angle of elevation changes to $30^{\circ}$. If the jet is flying at a speed of 648 kmph, find the constant height at which the jet is flying.
$J_1$: Initial position of the jet
$J_2$: Position of jet after 10 seconds $\left[\text{i.e. } \dfrac{10}{3600}=\dfrac{1}{360} \text{hr}\right]$
O: Point of observation
Speed of jet = 648 km/hr
$\therefore$ Distance covered by the jet in 10s $= \text{speed}\times \text{time}=648\times\dfrac{1}{360}=1.8$ km
$\therefore$ From the figure, $J_1J_2 = 1.8$ $\cdots$ (1)
Also from the figure, $J_1J_2 = QP$ $\cdots$ (2) and $QO=QP+PO$ $\cdots$ (3)
From triangle $J_1PO$,
$\dfrac{J_1P}{PO}=\tan 60^{\circ} = \sqrt{3}$
$\implies$ $PO=\dfrac{J_1P}{\sqrt{3}} = \dfrac{h}{\sqrt{3}}$ $\cdots$ (4)
From triangle $J_2QO$,
$\dfrac{J_2Q}{QO} = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$
$\therefore$ $QO = J_2Q \times \sqrt{3} = h\sqrt{3}$
i.e. $QP + PO = h\sqrt{3}$ $\cdots$ (5) [from equation (3)]
In view of equations (1), (2) and (4), equation (5) becomes
$1.8+\dfrac{h}{\sqrt{3}} = h \sqrt{3}$
i.e. $\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right)h= 1.8$
i.e. $\dfrac{2}{\sqrt{3}}h = 1.8$
$\implies$ $h=\dfrac{1.8\times \sqrt{3}}{2}=1.558$
$\therefore$ Height at which the jet is flying = 1.558 km