Question

angle of elevation of a jet from a point P on the ground is 60˚. after a flight of 15seconds, angle of elevation becomes half of the previous angle. If the jet is flying at a speed of 720km/h. Find the constant height at which jet is flying

Answer 1

Answer:

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/sec720×

18

5

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}

3

1

=

h

x

\mathrm{x}=\frac{h}{\sqrt{3}}x=

3

h

……………….. (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