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 = 36003/15 = 2403 km/hr.
Speed of aeroplane in km/hr is 2403.
Answered by | 29th Jan, 2010, 10:24: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number