The angle of elevation of a jetfighter from a point A on the ground is 600
.After a flight of 15
seconds, the angle of elevation changes to 300
. If the jet is flying at a speed of 720km/hr, find the
constant height at which the jet is flying.
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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 = = 200 m/sec
Time taken is 15 sec.
Distance BC = LM = 200 15 m = 3000 m
In ALB,AL/bl
cot 60° = AL/BL
……………….. (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
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