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(D) Let the height of the jet plane be $h \, m$. Let the point on the ground be $A$. From the first position, $	an 60^{\circ} = \frac{h}{y}$ $\Righta

Vedclass · Accepted Answer

$1200$

Vedclass · Answer

(D) Let the height of the jet plane be $h \, m$. Let the point on the ground be $A$. From the first position, $	an 60^{\circ} = \frac{h}{y}$ $\Rightarrow \sqrt{3} = \frac{h}{y}$ $\Rightarrow y = \frac{h}{\sqrt{3}} \quad \dots (1)$ After $20 \, s$, the plane covers a distance $x$. Speed $= 432 \, km/h = 432 	imes \frac{5}{18} \, m/s = 120 \, m/s$. Distance $x = 	ext{speed} 	imes 	ext{time} = 120 	imes 20 = 2400 \, m$. From the second position, $	an 30^{\circ} = \frac{h}{x + y}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{2400 + y}$ $\Rightarrow 2400 + y = h\sqrt{3} \quad \dots (2)$ Substitute $y$ from $(1)$ into $(2)$: $2400 + \frac{h}{\sqrt{3}} = h\sqrt{3}$ $2400 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} ight) = \frac{2h}{\sqrt{3}}$ $h = \frac{2400 	imes \sqrt{3}}{2} = 1200 \sqrt{3} \, m$.

The angle of elevation of a jet plane from a point $A$ on the ground is $60^{\circ}$. After a flight of $20 \, s$ at the speed of $432 \, km/h$, the angle of elevation changes to $30^{\circ}$. If the jet plane is flying at a constant height, then its height is ..... $m$. (in $\sqrt{3}$)

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