Question

The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 10 seconds the angle of elevation changes to 30°. If the jet is flying at a speed of 432km/hr, find the constant height at which the jet is flying.​

Answer 1

Answer:

Let O be the point of observation on the ground OX.

Let A and B be the two positions of the jet.

Then, /_ XOA = 60° and /_ XOB = 30°.

Draw Al perpendicular on OX and BM perpendicular on OX.

Let AL = BM = h metres.

Speed of the jet

= 432 × 5/ 18

= 24×5

= 120 m/s.

Time taken to cover the distance AB = 10 second

Distance covered = speed × time

= 120m/s × 10 s

= 1200 m

:. LM = AB = 1200m

Let OL = x metres

From right angle triangle OLA,

OL/ AL = COT 60°

x/h = 1/√3

x= h/√3 .........(1)

From right angle triangle OMB,

OM / BM = COT 30 °

( x+1200)/h = √3

x = (h√3 - 1200) ........(2)

Equating the value of x from (1) and (2)

h/√3 =(h√3 - 1200)

h = √3(h√3 - 1200)

h= 3h - 1200√3

2h= 1200√3

2h = 1200 × 1.732

2h = 2,078.4

h = 1,039.2

Hence, the required height is 1,039.2 m.