The angle of elevation of a Jet fighter from a point $A$ on the ground is $60^{o}$ . After $10$ seconds flight, the angle of elevation changes to $30^{o}$ . If the Jet is flying at a speed of $432 k m / h o u r$ , find the height at which the jet is flying.

2751.3 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1039.2 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3741.9 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1369.1 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1039.2 m$
Speed = 432 km.hr = $432 \times \frac{5}{18}$ = 120 m/sec
Time = 10 sec
Distance travelled = $120 \times 10$ = $1200$ m
Now, $tan ∠$ of elevation = $\frac{h e i g h t}{h o r i z o n t a l d i s t a n c e}$
Thus, $tan 60 = \frac{h}{d_{a}}$
$d_{a} = \frac{h}{\sqrt{3}}$
$tan 30 = \frac{h}{d_{b}}$
$d_{b} = h \sqrt{3}$
Thus, $d_{b} - d_{a} = d i s t a n c e t r a v e l l e d$
$h \sqrt{3} - \frac{h}{\sqrt{3}} = 1200$
$\frac{2 h}{\sqrt{3}} = 1200$
$h = 1039.2$ m

Suggest Corrections

Join BYJU'S Learning Program