Question

The angle of elevation of a jet fighter from a point A on the ground is 60 degree . after a flight of 15 seconds the angle of elevation changes to 30 degree .if the jet is flying at the speed 720 km/h find the constant height at which the jet is flying

Answer 1

H = 1500√3 m = 1500×1.732 = 2598 m

Answer 2

Answer:

The constant height at which the jet is flying is 2598 m

Step-by-step explanation:

Step 1:

Given data:

Let A be the point of observation on the ground and B and C be the two positions of the jet. Let BL=CM = h metres. Let AL = x metres.

Step 2:

Speed of the jet = 720 km/hr = $720 \times \frac{5}{18}m/sec$ = 200 m/sec

Time taken is 15 sec.

Distance BC = LM = 200   15 m = 3000 m

In  ALB,AL/bl

cot 60° = AL/BL

$\frac{1}{\sqrt{3}}=\frac{x}{h}$

$\mathrm{x}=\frac{h}{\sqrt{3}}$……………….. (i)

In  AMC,

Step 3:

cot 30° = AM/MC

=( AL+LM/MC )

x+3000 = h √3

x= h √3 -3000 ………………(ii)

Step 4:

From Equation (i) and Equation (ii)

√3 h-3000 =h/(√3)

3h-3000 √3 = h

2h= 5196

h=2598