Question

The angle of elevation of a jet plane form a point P on the ground is 60°. After a flight of 15 second, the angle of elevation changes to 30°. If the jet plane is flying at a constant of $1500\sqrt{3}\mathrm{m}$, then the jet plane is
(a) 360 km/hr
(b) 720 km/hr
(c) 540 km/hr
(d) 270 km/hr

Answer 1

(b) 720 km/h
Let $C$ be the position of the jet plane and $P$ be the point on the ground such that ∠$CPD={60}^o$, ∠$C\text{'}PD\text{'}={30}^o$ and $CD=1500\sqrt{3}$ m.
Let $DD\text{'}=x$ m and $PD=y$ m.

In the right ∆$CPD$, we have:
$\frac{CD}{PD}=\tan {60}^o=\sqrt{3}$
⇒ $\frac{1500\sqrt{3}}{y}=\sqrt{3}$
⇒ $y=1500$ m
Now, in the right ∆$C\text{'}PD\text{'}$, we have:
$\frac{C\text{'}D\text{'}}{D\text{'}P}=\tan {30}^o=\frac{1}{\sqrt{3}}$
⇒ $\frac{1500\sqrt{3}}{(x+y)}=\frac{1}{\sqrt{3}}$
⇒ $x+y=1500\sqrt{3}\times \sqrt{3}=4500$
Putting the value of $y=1500$ in the above equation, we get:
$x+1500=4500$ ⇒ $x=4500-1500=3000$ m
∴ Distance between the two points = $DD\text{'}=x=3000$ m
$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{3000}{15}=200$m/s = $200\times \frac{18}{5}=720$ km/h