plzzzz answer sir !!!!exam 2mrw!!!!

Asked by SINGALAA | 28th Jan, 2010, 07:37: PM

Expert Answer:

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 = 36003/15 = 2403 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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