plzzzz answer sir !!!!exam 2mrw!!!!
Asked by SINGALAA | 28th Jan, 2010, 07:37: PM
As the angle of elevation changes from 60 to 30, the aeroplane is moving away.
Let the speed be V (km/sec) and initial horizontal distance from the observe be X.
tan 60 = 1.5 km/X
X = 1.5/tan 60 = 1.5/3
After 15 sec, the distance increases by 15V.
tan 30 = 1.5/(X+15V)
(X+15V) = 1.5/tan30 = 1.53
15V = 1.53 - 1.5/
3 = 1.5(3-1)/
3 = 3/
3 =
3
V = 3/15 km/sec = 3600
3/15 = 240
3 km/hr.
Speed of aeroplane in km/hr is 2403.
Regards,
Team,
TopperLearning.
Answered by | 29th Jan, 2010, 10:24: PM
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